Bài tập toán cao cấp Tập 2 Nguyễn Thủy Thanh
NXB Đại học quốc gia Hà Nội 2007, 158 Tr.
Từ khoá: Bài tập toán cao cấp, Giới hạn dãy số, Giới hạn hàm số, Tính liên tục
của hàm số, Hàm liên tục, Phép tính vi phân hàm một biến,
Đạo hàm, Vi phân, Công thức Taylor, Đạo hàm riêng, Vi phân của hàm nhiều
biến, Cực trị của hàm nhiều biến.
Tài liệu trong Thư viện điện tử ĐH Khoa học Tự nhiên có thể sử dụng cho mục
đích học tập và nghiên cứu cá nhân. Nghiêm cấm mọi hình thức sao chép, in ấn
phục vụ các mục đích khác nếu không được sự chấp thuận của nhà xuất bản và
tác giả.
NGUYˆE˜ N THUY’ THANH
B `AI T ˆA. P TO ´AN CAO C ˆA´P
Tˆa.p 2 Ph´ep t´ınh vi phˆan c´ac h`am
NH `A XU ˆA´T BA’ N DA. I HO. C QU ˆO´C GIA H `A N ˆO. I
Mu. c lu. c
3 7 Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´ 4
5
. hˆo. i tu. cu’ a d˜ay sˆo´ du. 11
7.1 Gi´o.i ha. n cu’ a d˜ay sˆo´ . . . . . . . . . . . . . . . . . . . 7.1.1 C´ac b`ai to´an liˆen quan t´o.i di.nh ngh˜ıa gi´o.i ha. n . .a trˆen c´ac 7.1.2 Ch´u.ng minh su. di.nh l´y vˆe` gi´o.i ha. n . . . . . . . . . . . . . . . . .a trˆen diˆe`u . hˆo. i tu. cu’ a d˜ay sˆo´ du. 7.1.3 Ch´u.ng minh su.
17
7.1.4 Ch´u.ng minh su.
25
27
27
41
51 kiˆe.n du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y Bolzano-Weierstrass) . . . . . . . . . . . . . . . .a trˆen diˆe`u . hˆo. i tu. cu’ a d˜ay sˆo´ du. kiˆe.n cˆa` n v`a du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y hˆo. i tu. . . . . . . . . . . . . . . . . . Bolzano-Cauchy) 7.2 Gi´o.i ha. n h`am mˆo. t biˆe´n . . . . . . . . . . . . . . . . . . 7.2.1 C´ac kh´ai niˆe.m v`a di.nh l´y co. ba’ n vˆe` gi´o.i ha. n . . 7.3 H`am liˆen tu. c . . . . . . . . . . . . . . . . . . . . . . . 7.4 Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n . . . . . . . .
60 8 Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n
61
61
62
75
8.1 D- a. o h`am . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 D- a. o h`am cˆa´p 1 . . . . . . . . . . . . . . . . . . 8.1.2 D- a. o h`am cˆa´p cao . . . . . . . . . . . . . . . . . 8.2 Vi phˆan . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Vi phˆan cˆa´p 1 . . . . . . . . . . . . . . . . . . . 75
2 MU. C LU. C
77
84 84
(L’Hospitale)
88 96 8.2.2 Vi phˆan cˆa´p cao . . . . . . . . . . . . . . . . . . 8.3 C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi. Quy t˘a´c l’Hospital. Cˆong th´u.c Taylor . . . . . . . . . . . . . . . . . . . . . 8.3.1 C´ac d i.nh l´y co. ba’ n vˆe` h`am kha’ vi . . . . . . . . 8.3.2 Khu.’ c´ac da. ng vˆo di.nh. Quy t˘a´c Lˆopitan . . . . . . . . . . . . . . . . . . . 8.3.3 Cˆong th´u.c Taylor . . . . . . . . . . . . . . . . .
9 Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n
9.1 D- a. o h`am riˆeng
9.3 Cu.
109 . . . . . . . . . . . . . . . . . . . . . . 110 9.1.1 D- a. o h`am riˆeng cˆa´p 1 . . . . . . . . . . . . . . . 110 .p . . . . . . . . . . . . . . 111 9.1.2 D- a. o h`am cu’ a h`am ho. 9.1.3 H`am kha’ vi . . . . . . . . . . . . . . . . . . . . 111 9.1.4 D- a. o h`am theo hu.´o.ng . . . . . . . . . . . . . . . 112 9.1.5 D- a. o h`am riˆeng cˆa´p cao . . . . . . . . . . . . . . 113 9.2 Vi phˆan cu’ a h`am nhiˆe`u biˆe´n . . . . . . . . . . . . . . . 125 9.2.1 Vi phˆan cˆa´p 1 . . . . . . . . . . . . . . . . . . . 126 ´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung . . . . . . . 126 9.2.2 9.2.3 C´ac t´ınh chˆa´t cu’ a vi phˆan . . . . . . . . . . . . 127 9.2.4 Vi phˆan cˆa´p cao . . . . . . . . . . . . . . . . . . 127 9.2.5 Cˆong th´u.c Taylor . . . . . . . . . . . . . . . . . 129 9.2.6 Vi phˆan cu’ a h`am ˆa’n . . . . . . . . . . . . . . . 130 .c tri. cu’ a h`am nhiˆe`u biˆe´n . . . . . . . . . . . . . . . 145 .c tri. . . . . . . . . . . . . . . . . . . . . . . . 145 9.3.1 Cu. .c tri. c´o diˆe`u kiˆe.n . . . . . . . . . . . . . . . . 146 9.3.2 Cu. 9.3.3 Gi´a tri. l´o.n nhˆa´t v`a b´e nhˆa´t cu’ a h`am . . . . . . 147
Chu.o.ng 7
Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
4
5
7.1 Gi´o.i ha. n cu’ a d˜ay sˆo´ . . . . . . . . . . . . . . 7.1.1 C´ac b`ai to´an liˆen quan t´o.i di.nh ngh˜ıa gi´o.i ha. n . . . . . . . . . . . . . . . . . . . . . . .a trˆen
7.1.2 Ch´u.ng minh su.
. hˆo. i tu. cu’ a d˜ay sˆo´ du. c´ac di.nh l´y vˆe` gi´o.i ha. n . . . . . . . . . . . . 11
7.1.3 Ch´u.ng minh su.
.a . hˆo. i tu. cu’ a d˜ay sˆo´ du.
trˆen diˆe`u kiˆe.n du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y Bolzano-Weierstrass) . . . . . . . . 17
. hˆo. i tu. cu’ a d˜ay sˆo´ du.
7.1.4 Ch´u.ng minh su.
.a trˆen diˆe`u kiˆe.n cˆa` n v`a du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y hˆo. i tu. Bolzano-Cauchy) . . . . . . . . . . 25 7.2 Gi´o.i ha. n h`am mˆo.t biˆe´n . . . . . . . . . . . . 27 7.2.1 C´ac kh´ai niˆe.m v`a di.nh l´y co. ba’n vˆe` gi´o.i ha. n 27 7.3 H`am liˆen tu. c . . . . . . . . . . . . . . . . . . 41 7.4 Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n . 51
4 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
.p N du.o. .c go. i l`a d˜ay sˆo´ vˆo ha. n. D˜ay sˆo´
7.1 Gi´o.i ha. n cu’ a d˜ay sˆo´ H`am sˆo´ x´ac di.nh trˆen tˆa. p ho. .c viˆe´t du.´o.i da. ng: thu.`o.ng du.o.
(7.1) a1, a2, . . . , an, . . .
.c go. i l`a sˆo´ ha. ng tˆo’ng qu´at
ho˘a. c {an}, trong d´o an = f (n), n ∈ N du.o. cu’ a d˜ay, n l`a sˆo´ hiˆe.u cu’ a sˆo´ ha. ng trong d˜ay. Ta cˆa` n lu.u ´y c´ac kh´ai niˆe.m sau dˆay: i) D˜ay (7.1) du.o. .c go. i l`a bi. ch˘a. n nˆe´u ∃ M ∈ R+ : ∀ n ∈ N ⇒ |an| 6
M; v`a go. i l`a khˆong bi. ch˘a.n nˆe´u: ∀ M ∈ R+ : ∃ n ∈ N ⇒ |an| > M. .c go. i l`a gi´o.i ha. n cu’ a d˜ay (7.1) nˆe´u: ii) Sˆo´ a du.o.
(7.2) ∀ ε > 0, ∃ N (ε) : ∀ n > N ⇒ |an − a| < ε.
iii) Sˆo´ a khˆong pha’ i l`a gi´o.i ha. n cu’ a d˜ay (7.1) nˆe´u:
(7.3) ∃ ε > 0, ∀ N : ∃ n > N ⇒ |an − a| > ε.
.c go. i l`a d˜ay hˆo. i tu. , trong tru.`o.ng ho. .c .p ngu.o.
iv) D˜ay c´o gi´o.i ha. n du.o. la. i d˜ay (7.1) go. i l`a d˜ay phˆan k`y.
v) D˜ay (7.1) go. i l`a d˜ay vˆo c`ung b´e nˆe´u lim n→∞
an = 0 v`a go. i l`a d˜ay vˆo c`ung l´o.n nˆe´u ∀ A > 0, ∃ N sao cho ∀ n > N ⇒ |an| > A v`a viˆe´t lim an = ∞.
vi) Diˆe`u kiˆe.n cˆa` n dˆe’ d˜ay hˆo. i tu. l`a d˜ay d´o pha’ i bi. ch˘a. n. Ch´u ´y: i) Hˆe. th´u.c (7.2) tu.o.ng du.o.ng v´o.i:
(7.4) −ε < an − a < ε ⇔ a − ε < an < a + ε.
5 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
Hˆe. th´u.c (7.4) ch´u.ng to’ r˘a`ng mo. i sˆo´ ha. ng v´o.i chı’ sˆo´ n > N cu’ a d˜ay hˆo. i tu. dˆe`u n˘a`m trong khoa’ ng (a − ε, a + ε), khoa’ ng n`ay go. i l`a ε-lˆan cˆa. n cu’ a diˆe’m a.
Nhu. vˆa. y, nˆe´u d˜ay (7.1) hˆo. i tu. dˆe´n sˆo´ a th`ı mo. i sˆo´ ha. ng cu’ a n´o tr`u. ra mˆo. t sˆo´ h˜u.u ha. n sˆo´ ha. ng dˆe`u n˘a`m trong ε-lˆan cˆa. n bˆa´t k`y b´e bao nhiˆeu t`uy ´y cu’ a diˆe’m a.
7.1.1 C´ac b`ai to´an liˆen quan t´o.i di.nh ngh˜ıa gi´o.i
ha. n
ii) Ta lu.u ´y r˘a`ng d˜ay sˆo´ vˆo c`ung l´o.n khˆong hˆo. i tu. v`a k´y hiˆe.u lim an = ∞ (−∞) chı’ c´o ngh˜ıa l`a d˜ay an l`a vˆo c`ung l´o.n v`a k´y hiˆe.u d´o ho`an to`an khˆong c´o ngh˜ıa l`a d˜ay c´o gi´o.i ha. n.
.i) sao cho |an − a| 6 bn ∀ n v`a Dˆe’ ch´u.ng minh lim an = a b˘a`ng c´ach su.’ du. ng di.nh ngh˜ıa, ta cˆa` n tiˆe´n h`anh theo c´ac bu.´o.c sau dˆay: i) Lˆa. p biˆe’u th´u.c |an − a| ii) Cho. n d˜ay bn (nˆe´u diˆe`u d´o c´o lo. v´o.i ε du’ b´e bˆa´t k`y bˆa´t phu.o.ng tr`ınh dˆo´i v´o.i n:
(7.5) bn < ε
c´o thˆe’ gia’ i mˆo. t c´ach dˆe˜ d`ang. Gia’ su.’ (7.5) c´o nghiˆe.m l`a n > f (ε), f (ε) > 0. Khi d´o ta c´o thˆe’ lˆa´y n l`a [f (ε)], trong d´o [f (ε)] l`a phˆa` n nguyˆen cu’ a f (ε).
C ´AC V´I DU.
V´ı du. 1. Gia’ su.’ an = n(−1)n. Ch´u.ng minh r˘a`ng:
i) D˜ay an khˆong bi. ch˘a. n. ii) D˜ay an khˆong pha’ i l`a vˆo c`ung l´o.n. Gia’ i. i) Ta ch´u.ng minh r˘a`ng an tho’a m˜an di.nh ngh˜ıa d˜ay khˆong bi. ch˘a. n. Thˆa. t vˆa. y, ∀ M > 0 sˆo´ ha. ng v´o.i sˆo´ hiˆe.u n = 2([M] + 1) b˘a`ng n v`a l´o.n ho.n M. Diˆe`u d´o c´o ngh˜ıa l`a d˜ay an khˆong bi. ch˘a. n.
6 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
ii) Ta ch´u.ng minh r˘a`ng an khˆong pha’ i l`a vˆo c`ung l´o.n. Thˆa. t vˆa. y, ta x´et khoa’ ng (−2, 2). Hiˆe’n nhiˆen mo. i sˆo´ ha. ng cu’ a d˜ay v´o.i sˆo´ hiˆe.u le’ dˆe`u thuˆo. c khoa’ ng (−2, 2) v`ı khi n le’ th`ı ta c´o:
n(−1)n = n−1 = 1/n ∈ (−2, 2).
Nhu. vˆa. y trong kho’ng (−2, 2) c´o vˆo sˆo´ sˆo´ ha. ng cu’ a d˜ay. T`u. d´o,
theo di.nh ngh˜ıa suy ra an khˆong pha’ i l`a vˆo c`ung l´o.n. N V´ı du. 2. D`ung di.nh ngh˜ıa gi´o.i ha. n d˜ay sˆo´ dˆe’ ch´u.ng minh r˘a`ng:
1) = 0. 2) = 1. lim n→∞ lim n→∞ (−1)n−1 n n n + 1
Gia’ i. Dˆe’ ch´u.ng minh d˜ay an c´o gi´o.i ha. n l`a a, ta cˆa` n ch´u.ng minh r˘a`ng dˆo´i v´o.i mˆo˜i sˆo´ ε > 0 cho tru.´o.c c´o thˆe’ t`ım du.o. .c sˆo´ N (N phu. thuˆo. c ε) sao cho khi n > N th`ı suy ra |an − a| < ε. Thˆong thu.`o.ng ta c´o thˆe’ chı’ ra cˆong th´u.c tu.`o.ng minh biˆe’u diˆe˜n N qua ε.
1) Ta c´o:
· = (−1)n−1 n 1 n (cid:12) (cid:12) (cid:12) |an − 0| = (cid:12) (cid:12) (cid:12)
Gia’ su.’ ε l`a sˆo´ du.o.ng cho tru.´o.c t`uy ´y. Khi d´o:
· < ε ⇔ n > 1 n 1 ε
V`ı thˆe´ ta c´o thˆe’ lˆa´y N l`a sˆo´ tu.
⇒ N > < ε. . nhiˆen n`ao d´o tho’a m˜an diˆe`u kiˆe.n: 1 ε 1 N
(Ch˘a’ ng ha. n, ta c´o thˆe’ lˆa´y N = [1/ε], trong d´o [1/ε] l`a phˆa` n nguyˆen cu’ a 1/ε).
Khi d´o ∀ n > N th`ı:
6 < ε. |an − 0| = 1 n 1 N
7 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
= 0. (−1)n n . nhiˆen N (ε) sao cho ∀ n > Diˆe`u d´o c´o ngh˜ıa l`a lim n→∞ 2) Ta lˆa´y sˆo´ ε > 0 bˆa´t k`y v`a t`ım sˆo´ tu. N (ε) th`ı:
< ε. n n + 1 (cid:12) (cid:12) (cid:12) − 1(cid:12) (cid:12) (cid:12)
Bˆa´t d˘a’ ng th´u.c
< ε ⇔ − 1. |an − 1| < ε ⇔ 1 n + 1 1 ε
− 1, t´u.c l`a: Do d´o ta c´o thˆe’ lˆa´y sˆo´ N (ε) l`a phˆa` n nguyˆen cu’ a 1 ε
N (ε) = E((1/ε) − 1).
Khi d´o v´o.i mo. i n > N ta c´o:
6 = = 1. N < ε ⇒ lim n→∞ n n + 1 1 N + 1 1 n + 1 (cid:12) (cid:12) (cid:12) − 1(cid:12) (cid:12) (cid:12)
n n + 1 V´ı du. 3. Ch´u.ng minh r˘a`ng c´ac d˜ay sau dˆay phˆan k`y:
(7.6) 1)
(7.7) 2) an = n, n ∈ N an = (−1)n, n ∈ N
· (7.8) 3) an = (−1)n +
1 n Gia’ i. 1) Gia’ su.’ d˜ay (7.6) hˆo. i tu. v`a c´o gi´o.i ha. n l`a a. Ta lˆa´y ε = 1. Khi d´o theo di.nh ngh˜ıa gi´o.i ha. n tˆo` n ta. i sˆo´ hiˆe.u N sao cho ∀ n > N th`ı ta c´o |an − a| < 1 ngh˜ıa l`a |n − a| < 1 ∀ n > N . T`u. d´o −1 < n − a < 1 ∀ n > N ⇔ a − 1 < n < a + 1 ∀ n > N .
.p c´ac
sˆo´ tu.
Nhu.ng bˆa´t d˘a’ ng th´u.c n < a + 1, ∀ n > N l`a vˆo l´y v`ı tˆa. p ho. . nhiˆen khˆong bi. ch˘a. n. 2) C´ach 1. Gia’ su.’ d˜ay an hˆo. i tu. v`a c´o gi´o.i ha. n l`a a. Ta lˆa´y lˆan
, a + cˆa. n (cid:16)a − (cid:17) cu’ a diˆe’m a. Ta viˆe´t d˜ay d˜a cho du.´o.i da. ng: 1 2 1 2
(7.9) {an} = −1, 1, −1, 1, . . . .
8 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
, a + V`ı dˆo. d`ai cu’ a khoa’ ng (cid:16)a − 1 2 1 2
, a + (cid:17) l`a b˘a`ng 1 nˆen hai diˆe’m −1 1 2 1 2
, a + 1 2 v`a +1 khˆong thˆe’ dˆo` ng th`o.i thuˆo. c lˆan cˆa. n (cid:16)a − (cid:17) cu’ a diˆe’m a, v`ı khoa’ ng c´ach gi˜u.a −1 v`a +1 b˘a`ng 2. Diˆe`u d´o c´o ngh˜ıa l`a o.’ ngo`ai 1 (cid:17) c´o vˆo sˆo´ sˆo´ ha. ng cu’ a d˜ay v`a v`ı thˆe´ (xem ch´u lˆan cˆa. n (cid:16)a − 2 ´y o.’ trˆen) sˆo´ a khˆong thˆe’ l`a gi´o.i ha. n cu’ a d˜ay.
) ta c´o C´ach 2. Gia’ su.’ an → a. Khi d´o ∀ ε > 0 (lˆa´y ε = 1 2
∀ n > N. |an − a| < 1 2
V`ı an = ±1 nˆen
| − 1 − a| < , |1 − a| < 1 2 1 2
+ = 1 ⇒2 = |(1 − a) + (1 + a)| 6 |1 − a| + |a + 1| 6 1 2 1 2
⇒2 < 1, vˆo l´y.
3) Lu.u ´y r˘a`ng v´o.i n = 2m ⇒ a2m = 1 + . Sˆo´ ha. ng kˆe` v´o.i n´o 1 2m c´o sˆo´ hiˆe.u le’ 2m + 1 (hay 2m − 1) v`a
< 0 6 0). a2m+1 = −1 + (hay a2m−1 = −1 + 1 2m + 1 1 2m − 1
T`u. d´o suy r˘a`ng
|an − an−1| > 1.
. Khi d´o Nˆe´u sˆo´ a n`ao d´o l`a gi´o.i ha. n cu’ a d˜ay (an) th`ı b˘a´t dˆa` u t`u. sˆo´ hiˆe.u n`ao d´o (an) tho’a m˜an bˆa´t d˘a’ ng th´u.c |an − a| < 1 2
+ = 1. |an − an+1| 6 |an − a| + |an+1 − a| < 1 2 1 2
Nhu.ng hiˆe.u gi˜u.a hai sˆo´ ha. ng kˆe` nhau bˆa´t k`y cu’ a d˜ay d˜a cho luˆon luˆon l´o.n ho.n 1. Diˆe`u mˆau thuˆa˜n n`ay ch´u.ng to’ r˘a`ng khˆong mˆo. t sˆo´ thu. .c n`ao c´o thˆe’ l`a gi´o.i ha. n cu’ a d˜ay d˜a cho. N
9 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
B `AI T ˆA. P
H˜ay su.’ du. ng di.nh ngh˜ıa gi´o.i ha. n dˆe’ ch´u.ng minh r˘a`ng
1. an = 1 nˆe´u an = lim n→∞
2. an = 3 5 2n − 1 2n + 2 3n2 + 1 nˆe´u an = lim 5n2 − 1 n→∞ B˘a´t dˆa` u t`u. sˆo´ hiˆe.u N n`ao th`ı:
(DS. N = 5) |an − 3/5| < 0, 01
. 3. an = 1 nˆe´u an = lim n→∞ 3n + 1 3n
4. = 0. lim n→∞ cos n n
3
= 5. 5. lim n→∞ 2n + 5 · 6n 3n + 6n √
= 0. 6. lim n→∞ n2 sin n2 n + 1
.
7. Ch´u.ng minh r˘a`ng sˆo´ a = 0 khˆong pha’ i l`a gi´o.i ha. n cu’ a d˜ay an = n2 − 2 2n2 − 9 8. Ch´u.ng minh r˘a`ng
= 1. lim n→∞ n2 + 2n + 1 + sin n n2 + n + 1
, . . .
2
9. Ch´u.ng minh r˘a`ng d˜ay: an = (−1)n + 1/n phˆan k`y. 10. Ch´u.ng minh r˘a`ng d˜ay; an = sin n0 phˆan k`y. 11. T`ım gi´o.i ha. n cu’ a d˜ay: 0, 2; 0, 22; 0, 222; . . . , 0, 22 . . . 2 | {z }n Chı’ dˆa˜n. Biˆe’u diˆe˜n an du.´o.i da. ng
+ + · · · + an = 0, 22 . . . 2 = (DS. lim an = 2/9) 2 10 2 10 2 10n
10 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
cu’ a d˜ay sˆo´: gi´o.i T`ım
ha. n , . . .
12. 0, 2; 0, 23; 0, 233; 0, 2333; . . . , 0, 2 33 . . . 3 | {z }n Chı’ dˆa˜n. Biˆe’u diˆe˜n an du.´o.i da. ng
+ (cid:16) + + · · · + (cid:17) (DS. 7/30) an = 2 10 3 102 3 103 3 10n
= 0. i) lim n→∞
= 0 (a > 1). lim n→∞ 13. Ch´u.ng minh r˘a`ng nˆe´u d˜ay an hˆo. i tu. dˆe´n a, c`on d˜ay bn dˆa` n dˆe´n ∞ th`ı d˜ay an/bn dˆa` n dˆe´n 0. 14. Ch´u.ng minh r˘a`ng n 2n n an
· + · · · + 1 > n + > 2n = (1 + 1)n = 1 + n + ii) Chı’ dˆa˜n. i) Su.’ du. ng hˆe. th´u.c: n(n − 1) 2 n(n − 1) 2 n2 2
.ng |an − 0|. v`a u.´o.c lu.o.
ii) Tu.o.ng tu. . nhu. i). Su.’ du. ng hˆe. th´u.c:
(a − 1). an = [1 + (a − 1)]n > n(n − 1) 2
15. Ch´u.ng minh r˘a`ng
+ · · · + lim an = 2 nˆe´u an = 1 + 1 2 1 2n
Chı’ dˆa˜n. ´Ap du. ng cˆong th´u.c t´ınh tˆo’ng cˆa´p sˆo´ nhˆan dˆe’ t´ınh an rˆo` i
.ng |an − 2|.
u.´o.c lu.o. 16. Biˆe´t r˘a`ng d˜ay an c´o gi´o.i ha. n, c`on d˜ay bn khˆong c´o gi´o.i ha. n. C´o thˆe’ n´oi g`ı vˆe` gi´o.i ha. n cu’ a d˜ay:
i) {an + bn}. ii) {anbn}. (DS. i) lim{an + bn} khˆong tˆo` n ta. i. H˜ay ch´u.ng minh.
11 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
.p c´o gi´o.i ha. n v`a khˆong c´o gi´o.i ha. n, ii) C´o thˆe’ g˘a. p ca’ hai tru.`o.ng ho.
v´ı du. :
.a trˆen
an = , bn = (−1)n; an = , bn = (−1)n. n − 1 n
7.1.2 Ch´u.ng minh su.
c´ac di.nh l´y vˆe` gi´o.i ha. n
1 n . hˆo. i tu. cu’ a d˜ay sˆo´ du.
Dˆe’ t´ınh gi´o.i ha. n cu’ a d˜ay sˆo´, ngu.`o.i ta thu.`o.ng su.’ du. ng c´ac di.nh l´y v`a kh´ai niˆe.m sau dˆay:
Gia’ su.’ lim an = a, lim bn = b. i) lim(an ± bn) = lim an ± lim bn = a ± b. ii) lim anbn = lim an · lim bn = a · b. iii) Nˆe´u b 6= 0 th`ı b˘a´t dˆa` u t`u. mˆo. t sˆo´ hiˆe.u n`ao d´o d˜ay an/bn x´ac
di.nh (ngh˜ıa l`a ∃ N : ∀ n > N ⇒ bn 6= 0) v`a:
· = = lim a b an bn
lim an lim bn iv) Nˆe´u lim an = a, lim bn = a v`a b˘a´t dˆa` u t`u. mˆo. t sˆo´ hiˆe.u n`ao d´o
an 6 zn 6 bn th`ı lim zn = a (Nguyˆen l´y bi. ch˘a. n hai phi´a).
(cid:17) l`a d˜ay vˆo v) T´ıch cu’ a d˜ay vˆo c`ung b´e v´o.i d˜ay bi. ch˘a. n l`a d˜ay vˆo c`ung b´e. vi) Nˆe´u (an) l`a d˜ay vˆo c`ung l´o.n v`a an 6= 0 th`ı d˜ay (cid:16) 1 an
(cid:17) .c la. i, nˆe´u αn l`a d˜ay vˆo c`ung b´e v`a αn 6= 0 th`ı d˜ay (cid:16) 1 αn c`ung b´e; ngu.o. l`a vˆo c`ung l´o.n.
Nhˆa. n x´et. Dˆe’ ´ap du. ng d´ung d˘a´n c´ac di.nh l´y trˆen ta cˆa` n lu.u ´y mˆo. t
sˆo´ nhˆa. n x´et sau dˆay:
.c nˆe´u i) Di.nh l´y (iii) vˆe` gi´o.i ha. n cu’ a thu.o.ng s˜e khˆong ´ap du. ng du.o. tu.’ sˆo´ v`a mˆa˜u sˆo´ khˆong c´o gi´o.i ha. n h˜u.u ha. n ho˘a. c mˆa˜u sˆo´ c´o gi´o.i ha. n .p d´o nˆen biˆe´n dˆo’i so. bˆo. d˜ay thu.o.ng, b˘a`ng 0. Trong nh˜u.ng tru.`o.ng ho. ch˘a’ ng ha. n b˘a`ng c´ach chia ho˘a. c nhˆan tu.’ sˆo´ v`a mˆa˜u sˆo´ v´o.i c`ung mˆo. t biˆe’u th´u.c.
12 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
ii) Dˆo´i v´o.i di.nh l´y (i) v`a (ii) c˜ung cˆa` n pha’ i thˆa. n tro. ng khi ´ap du. ng. .p n`ay ta cˆa` n pha’ i biˆe´n dˆo’i c´ac biˆe’u th´u.c an ± bn v`a
Trong tru.`o.ng ho. an · bn tru.´o.c khi t´ınh gi´o.i ha. n (xem v´ı du. 1, iii). an = a. iii) Nˆe´u an = a ≡ const ∀ n th`ı lim n→∞
C ´AC V´I DU.
V´ı du. 1. T`ım lim an nˆe´u:
1) an = (1 + 7n+2)/(3 − 7n) 2) an = (2 + 4 + 6 + · · · + 2n)/[1 + 3 + 5 + · · · + (2n + 1)] 3) an = n3/(12 + 22 + · · · + n2) Gia’ i. Dˆe’ gia’ i c´ac b`ai to´an n`ay ta d`ung l´y thuyˆe´t cˆa´p sˆo´ 1) Nhˆan tu.’ sˆo´ v`a mˆa˜u sˆo´ phˆan th´u.c v´o.i 7−n ta c´o:
an = 1 + 7n+2 3 − 7n = 7−n + 72 3 · 7−n − 1
Do d´o
= −49 v`ı lim 7−n = 0, n → ∞. lim an = lim 7−n + 72 3 · 7−n − 1
2) Tu.’ sˆo´ v`a mˆa˜u sˆo´ dˆe`u l`a cˆa´p sˆo´ cˆo. ng nˆen ta c´o:
· n; 2 + 4 + 6 + · · · + 2n = 2 + 2n 2
(n + 1). 1 + 3 + 5 + · · · + (2n + 1) = 1 + (2n + 2) 2
Do d´o
an = ⇒ lim an = 1. n n + 1
3) Nhu. ta biˆe´t:
12 + 22 + · · · + n2 = n(n + 1)(2n + 1) 6
13 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
v`a do d´o:
lim an = lim
= lim = 3. N 6n3 n(n + 1)(2n + 1) 6 (1 + 1/n)(2 + 1/n)
V´ı du. 2. T`ım gi´o.i ha. n
+ + · · · + 1 + lim + + · · · + 1 + 1 2 1 3 1 4 1 9 1 2n 1 3n
Gia’ i. Tu.’ sˆo´ v`a mˆa˜u sˆo´ dˆe`u l`a cˆa´p sˆo´ nhˆan nˆen
+ · · · + , 1 +
+ · · · + 1 + 1 2 1 3 1 2n = 1 = 3n 2(2n − 1) 2n 3(3n − 1) 2 · 3n
v`a do d´o:
· · = 2 lim lim lim an = lim 2(2n − 1) 2n 2n − 1 2n 2 3
lim · 1 = · N = 2 lim[1 − (1/2)n] · = 2 · 1 · 2 · 3n 3(3n − 1) 2 3 1 1 − (1/3)n 3n 3n − 1 4 3 2 3
V´ı du. 3.
n2 + n − n √ n + 2 − 3 n n2 − n3 + n
√ √ ( √ 1) an = √ 2) an = 3 √ 3) an = 3 Gia’ i. 1) Ta biˆe´n dˆo’i an b˘a`ng c´ach nhˆan v`a chia cho da. i lu.o. n2 + n + n) n2 + n − n)( √ √ = = an = n n2 + n + n n2 + n + n .p .ng liˆen ho. 1 p1 + 1/n + 1
Do d´o
1 · = lim an = 1 2 (p1 + 1/n + 1) lim n→∞
14 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
. nhu. 1) ta c´o: 2) Biˆe´n dˆo’i an tu.o.ng tu.
3 n(cid:1) √ n + (cid:0) 3
2 n(cid:1)
2 n + 2(cid:1)
2 n(cid:1)
an = √ 3 − (cid:0) 3 √ n + 2 · 3 n + 2(cid:1) √ + 3 √ (cid:0) 3 √ (cid:0) 3 2 n + 2(cid:1) 2 an = √ + 3 √ n + 2 · 3 √ (cid:0) 3 √ n + (cid:0) 3
2 + 3p1 + 2/n + 1(cid:3) → ∞
Biˆe’u th´u.c mˆa˜u sˆo´ b˘a`ng:
n2/3(cid:2)(cid:0) 3p1 + 2/n(cid:1)
khi n → ∞ v`a do d´o lim an = 0. √ 3) Ta c´o thˆe’ viˆe´t n = 3 n3 v`a ´ap du. ng cˆong th´u.c:
a3 + b3 = (a + b)(a2 − ab + b2)
suy ra
√ (cid:0) 3 n2 − n3 + n2(cid:3) an = √ 2 − n 3 n2 − n3 + n2 n2 − n3(cid:1) √ 2 − n 3
= n2 − n3 + n2 n2 − n3(cid:1) √ (cid:0) 3
= √ n2 − n3 + n(cid:1)(cid:2)(cid:0) 3 √ n2 − n3(cid:1) (cid:0) 3 n2 √ 2 − n 3 1 [1/n − 1]2/3 − [1/n − 1]1/3 + 1
1 3
√ √ , , an = bn = · N suy ra lim an = V´ı du. 4. T`ım gi´o.i ha. n cu’ a c´ac d˜ay sau n n2 + 1
√ √ √ · + · · · + + cn = n n2 + n 1 n + 1 1 n2 + 2
= lim = 1. lim an = lim 1 n2 + n Gia’ i. Dˆa` u tiˆen ta ch´u.ng minh lim an = 1. Thˆa. t vˆa. y: n np1 + 1/n 1 p1 + 1/n
15 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
. lim bn = 1. Tu.o.ng tu.
Dˆe’ t`ım gi´o.i ha. n cu’ a cn ta s˜e ´ap du. ng Nguyˆen l´y bi. ch˘a.n hai ph´ıa.
Mˆo. t m˘a. t ta c´o:
√ √ √ √ + + · · · + = cn < = bn 1 n2 + 1 1 n2 + 1 n n2 + 1
1 n2 + 1 nhu.ng m˘a. t kh´ac:
√ √ √ + + · · · + cn > = an. 1 n2 + n 1 n2 + n
n→∞
1 n2 + n bn = 1. T`u. d´o suy ra an = lim n→∞ cn = 1. N Nhu. vˆa. y an < cn < bn v`a lim lim n→∞
V´ı du. 5. Ch´u.ng minh r˘a`ng d˜ay (qn) l`a: 1) d˜ay vˆo c`ung l´o.n nˆe´u |q| > 1; 2) d˜ay vˆo c`ung b´e khi |q| < 1.
n (cid:17)
Gia’ i. 1) Gia’ su.’ |q| > 1. Ta lˆa´y sˆo´ A > 0 bˆa´t k`y. T`u. d˘a’ ng th´u.c .c n > log|q|A. Nˆe´u ta lˆa´y N = [log|q|A] th`ı ∀ n > N
−1
> 1 nˆen |q|n > A ta thu du.o. ta c´o |q|n > A. Do d´o d˜ay (qn) l`a d˜ay vˆo c`ung l´o.n. −1 2) Gia’ su.’ |q| < 1, q 6= 0. Khi d´o qn = h(cid:16) i 1 q 1 q (cid:12) (cid:12) (cid:12)
ni (cid:1)
n(cid:17) l`a d˜ay vˆo c`ung l´o.n v`a do d´o d˜ay (cid:16)h(cid:0) d˜ay (cid:16)(cid:0) (cid:1) b´e, t´u.c l`a d˜ay (qn) l`a d˜ay vˆo c`ung b´e khi |q| < 1.
. V`ı (cid:12) (cid:12) (cid:12) (cid:17) l`a vˆo c`ung 1 q 1 q
3) Nˆe´u q = 0 th`ı qn = 0, |q|n < ε ∀ n v`a do d´o (qn) l`a vˆo c`ung b´e.
N
B `AI T ˆA. P
n→∞
an nˆe´u T`ım gi´o.i ha. n lim
. (DS. ∞) 1. an =
n2 − n √ n − n √ n2 + 1). (DS. −∞) 2. an = n2(n −
16 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
√ . (DS. 1/6) 3. an = 1 + 2 + 3 + · · · + n 9n4 + 1 √
. (DS. 0) 4. an =
+ . (DS. 5) 5. an =
− . (DS. 1/3) 6. an =
− . (DS. 1) 7. an = sin n n 3n2 3n + 1 cos n 10n
(DS. ∞) 8. an =
− . ) (DS. − 9. an = 3n 6n + 1 1 2
. (DS. 0) 10. an = n + 1
3
n cos n n + 1 5n n + 1 n3 n2 + 1 n n + 11 n3 + 1 n2 − 1 cos n3 n (−1)n √ 5 √ √ . (DS. +∞) √ √ 11. an = n2 + 1 + n3 + n − n n
1 − n3 + n. (DS. 0) √ 12. an = 3
3
√ √ (DS. 1) . 13. an = n2 + 4n n3 − 3n2
. (DS. −∞) 14. an =
− 2. (DS. −1) 15. an = (n + 3)! 2(n + 1)! − (n + 2)! 2 + 4 + · · · + 2n n + 2
) n3 − n2. (DS. √ 16. an = n − 3 1 3 1 − 2 + 3 − 4 + 5 − · · · − 2n √ √ . ) (DS. − 17. an = 1 3
+ . + · · · + 18. an = n2 + 1 + 1 2 · 3 1 1 · 2
− = (DS. 1) Chı’ dˆa˜n. ´Ap du. ng 4n2 + 1 1 n(n + 1) 1 n 1 n(n + 1) 1 n + 1
17 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
− + + · · · + ) . (DS. 19. an = 1 − 1 9 1 27 (−1)n−1 3n−1 3 4
. (DS. 3) 20. an =
. (DS. 1) 21. an = 1 3 2n+1 + 3n+1 2n + 3n n + (−1)n n − (−1)n
(cid:17)(cid:16) √ √ √ √ √ √ + + · · · + 22. an = (cid:16) 1 √ n 1 2n − 1 + 3 5 1 3 + 1 1 +
2n + 1 Chı’ dˆa˜n. Tru. c c˘an th´u.c o.’ mˆa˜u sˆo´ c´ac biˆe’u th´u.c trong dˆa´u ngo˘a. c.
) (DS.
+ + · · · + 23. an = 1 √ 2 1 1 · 2 · 3 1 n(n + 1)(n + 2)
h i − = ) (DS. 1 4
) + · · · + (DS. + . 24. an = 1 2 · 3 · 4 Chı’ dˆa˜n. Tru.´o.c hˆe´t ta ch´u.ng minh r˘a`ng 1 1 (n + 1)(n + 2) 2 1 a1d 1 n(n + 1)(n + 2) 1 1 a2a3 a1a2 1 n(n + 1) 1 anan+1
) (DS. 25. an = (1 − 1/4)(1 − 1/9) · · · (1 − 1/(n + 1)2).
.a trˆen
7.1.3 Ch´u.ng minh su.
. hˆo. i tu. cu’ a d˜ay sˆo´ du. diˆe`u kiˆe.n du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y Bolzano-Weierstrass)
. Chı’ dˆa˜n. B˘a`ng quy na. p to´an ho. c ch´u.ng to’ r˘a`ng an = trong d´o {an} l`a cˆa´p sˆo´ cˆo. ng v´o.i cˆong sai d 6= 0, an 6= 0. 1 2 n + 2 2n + 2
D˜ay sˆo´ an du.o.
.c go. i l`a: i) D˜ay t˘ang nˆe´u an+1 > an ∀ n ii) D˜ay gia’ m nˆe´u an+1 < an ∀ n C´ac d˜ay t˘ang ho˘a. c gia’ m c`on du.o.
.c go. i l`a d˜ay do.n diˆe.u. Ta lu.u ´y r˘a`ng d˜ay do.n diˆe.u bao gi`o. c˜ung bi. ch˘a. n ´ıt nhˆa´t l`a mˆo. t ph´ıa. Nˆe´u d˜ay
18 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Di.nh l´y n`ay kh˘a’ ng di.nh vˆe` su.
do.n diˆe.u t˘ang th`ı n´o bi. ch˘a.n du.´o.i bo.’ i sˆo´ ha. ng dˆa` u tiˆen cu’ a n´o, d˜ay do.n diˆe.u gia’ m th`ı bi. ch˘a. n trˆen bo.’ i sˆo´ ha. ng dˆa` u. Ta c´o di.nh l´y sau dˆay .c su.’ du. ng dˆe’ t´ınh gi´o.i ha. n cu’ a d˜ay do.n diˆe.u. thu.`o.ng du.o. D- i.nh l´y Bolzano-Weierstrass. D˜ay do.n diˆe. u v`a bi. ch˘a. n th`ı hˆo. i tu. . . tˆo` n ta. i cu’ a gi´o.i ha. n m`a khˆong chı’ .c phu.o.ng ph´ap t`ım gi´o.i ha. n d´o. Tuy vˆa. y, trong nhiˆe`u tru.`o.ng .p khi biˆe´t gi´o.i ha. n cu’ a d˜ay tˆo` n ta. i, c´o thˆe’ chı’ ra phu.o.ng ph´ap t´ınh .a trˆen d˘a’ ng th´u.c d´ung v´o.i mo. i d˜ay hˆo. i
ra du.o. ho. n´o. Viˆe.c t´ınh to´an thu.`o.ng du. tu. :
an. lim n→∞ an+1 = lim n→∞
.i ho.n ca’ l`a su.’ .a trˆen d˘a’ ng th´u.c v`u.a nˆeu tiˆe.n lo.
Khi t´ınh gi´o.i ha. n du. du. ng c´ach cho d˜ay b˘a`ng cˆong th´u.c truy hˆo` i.
C ´AC V´I DU.
V´ı du. 1. Ch´u.nh minh r˘a`ng d˜ay:
+ + · · · + an = hˆo. i tu. . 1 5 + 1 1 52 + 1 1 5n + 1
Gia’ i. D˜ay d˜a cho do.n diˆe.u t˘ang. Thˆa. t vˆa. y v`ı:
an+1 = an + nˆen an+1 > an. 1 5n+1 + 1
D˜ay d˜a cho bi. ch˘a. n trˆen. Thˆa. t vˆa. y:
+ + + · · · + < + an = 1 52 + 1 1 53 + 1 1 5n + 1 1 5 1 52 + · · · + 1 5n
− 1 5 + 1 1 5 · = (cid:16)1 − = (cid:17) < 1 5n 1 4 1 4 1 − 1 5n+1 1 5
Nhu. vˆa. y d˜ay an d˜a cho do.n diˆe.u t˘ang v`a bi. ch˘a. n trˆen nˆen n´o hˆo. i
tu. . N
19 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
hˆo. i tu. v`a t`ım gi´o.i ha. n cu’ a 2n n! V´ı du. 2. Ch´u.ng minh r˘a`ng d˜ay an = n´o.
, , . . . , , . . . 2 1 2n n! 22 Gia’ i. D˜ay d˜a cho c´o da. ng 2 D˜ay an do.n diˆe.u gia’ m. Thˆa. t vˆa. y
: = < 1 ∀ n > 1. = 2n+1 (n + 1)! 2n n! 2 n + 1 an+1 an
Do d´o an+1 < an v`a d˜ay bi. ch˘a.n trˆen bo.’ i phˆa` n tu.’ a1. Ngo`ai ra an > 0, ∀ n nˆen d˜ay bi. ch˘a.n du.´o.i. Do d´o d˜ay do.n diˆe.u gia’ m v`a bi. ch˘a.n. N´o hˆo. i tu. theo di.nh l´y Weierstrass. Gia’ su.’ a l`a gi´o.i ha. n cu’ a n´o. Ta c´o:
= ⇒ an+1 = an. 2 n + 1 2 n + 1 an+1 an
T`u. d´o
= lim lim an+1 = lim lim an 2an n + 1 2 n + 1
= 0. N v`a nhu. vˆa. y: a = 0 · a → a = 0. Vˆa. y: lim 2n n! √ √ 2, an+1 = 2an. Ch´u.ng minh r˘a`ng d˜ay hˆo. i
V´ı du. 3. Cho d˜ay an = tu. v`a t`ım gi´o.i ha. n cu’ a n´o.
√ Gia’ i. Hiˆe’n nhiˆen r˘a`ng: a1 < a2 < a3 < · · · < . D´o l`a d˜ay do.n diˆe.u 2. Ta ch´u.ng minh r˘a`ng n´o bi. ch˘a. n trˆen t˘ang v`a bi. ch˘a.n du.´o.i bo.’ i sˆo´ bo.’ i sˆo´ 2.
Thˆa. t vˆa. y
√ √ √ 2 · 2 = 2. a1 = 2; a2 = 2a1 <
.c r˘a`ng an 6 2.
Gia’ su.’ d˜a ch´u.ng minh du.o. Khi d´o:
√ √ 2 · 2 = 2. an+1 = 2an 6
20 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Vˆa.y theo tiˆen dˆe` quy na. p ta c´o an 6 2 ∀ n. Nhu. thˆe´ d˜ay an do.n diˆe.u t˘ang v`a bi. ch˘a. n nˆen n´o c´o gi´o.i ha. n d´o l`a a.
Ta c´o:
n+1 = 2an.
√ an+1 = 2an ⇒ a2
Do d´o:
n+1 = 2 lim an
lim a2
.c a1 = 0, a2 = 2. hay a2 − 2a = 0 v`a thu du.o.
V`ı d˜ay do.n diˆe.u t˘ang ∀ n nˆen gi´o.i ha. n a = 2. N
V´ı du. 4. Ch´u.ng minh t´ınh hˆo. i tu. v`a t`ım gi´o.i ha. n cu’ a d˜ay
√ √ a, . . . , x1 = a; x2 = qa +
√ r a + qa + · · · + a, a > 0, n dˆa´u c˘an. xn =
Gia’ i. i) R˜o r`ang: x1 < x2 < x3 < · · · < xn < xn+1 < . . . ngh˜ıa l`a
d˜ay d˜a cho l`a d˜ay t˘ang.
ii) Ta ch´u.ng minh d˜ay xn l`a d˜ay bi. ch˘a.n. Thˆa. t vˆa. y, ta c´o:
√ √ a < a + 1 x1 = √ √ √ √ a < qa + a + 1 < qa + 2 a + 1 = a + 1. x2 = qa +
√ a + 1. .c r˘a`ng: xn < √ Gia’ su.’ d˜a ch´u.ng minh du.o. Ta cˆa` n ch´u.ng minh xn+1 < a + 1. Thˆa. t vˆa. y, ta c´o:
√ √ √ √ a + 1 < qa + 2 a + 1 = a + 1. xn+1 = a + xn < qa +
Do d´o nh`o. ph´ep quy na. p to´an ho. c ta d˜a ch´u.ng minh r˘a`ng d˜ay d˜a √ a + 1. cho bi. ch˘a. n trˆen bo.’ i
21 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
√ a + xn−1 hay iii) Dˆe’ t`ım gi´o.i ha. n ta x´et hˆe. th´u.c xn =
x2 n = a + xn−1.
T`u. d´o:
n = lim(a + xn−1) = a + lim xn−1
lim x2
hay nˆe´u gia’ thiˆe´t lim xn = A th`ı: A2 = a + A → A2 − A − a = 0 v`a √ √ 1 + 1 − · A1 = , A2 = 1 + 4a 2
1 + 4a 2 V`ı A2 < 0 nˆen gi´a tri. A2 bi. loa. i v`ı xn > 0. Do d´o; √ 1 + · N lim xn = 1 + 4a 2
.c x´ac di.nh nhu. sau: a1 l`a sˆo´ V´ı du. 5. T`ım gi´o.i ha. n cu’ a d˜ay an du.o. t`uy ´y m`a
(7.10) 0 < a1 < 1, an+1 = an(2 − an) ∀ n > 1.
Gia’ i. i) Dˆa` u tiˆen ch´u.ng minh r˘a`ng an bi. ch˘a. n, m`a cu. thˆe’ l`a b˘a`ng
ph´ep quy na. p to´an ho. c ta ch´u.ng minh r˘a`ng
(7.11) 0 < an < 1.
.c ch´u.ng minh v´o.i n v`a ta Ta c´o 0 < a1 < 1. Gia’ su.’ (7.11) d˜a du.o. s˜e ch´u.ng minh (7.11) d´ung v´o.i n + 1 .
T`u. (7.10) ta c´o; an+1 = 1 − (1 − an)2. T`u. hˆe. th´u.c n`ay suy ra 0 < (1 − an)2 < 1, v`ı 0 < an < 1. T`u. d´o suy ra: 0 < an+1 < 1 ∀ n. ii) Bˆay gi`o. ta ch´u.ng minh r˘a`ng an l`a d˜ay t˘ang. Thˆa. t vˆa. y, v`ı an < 1 nˆen 2 − an > 1. Chia (7.10) cho an ta thu .c: du.o.
= 2 − an > 1. an+1 an
22 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
T`u. d´o an+1 > an ∀ n. Nhu. vˆa. y d˜ay an do.n diˆe.u t˘ang v`a bi. ch˘a. n. Do d´o theo di.nh l´y Weierstrass, lim An tˆo` n ta. i v`a ta k´y hiˆe.u n´o l`a a. iii) T`u. (7.10) ta c´o:
lim an+1 = lim an · lim(2 − an)
hay a = a(2 − a).
T`u. d´o a = 0 v`a a = 1. V`ı x1 > 0 v`a d˜ay an t˘ang nˆen
a = 1 = lim an. N
hˆo. i tu. v`a t`ım gi´o.i ha. n cu’ a n! nn V´ı du. 6. Ch´u.ng minh r˘a`ng d˜ay an = n´o.
Gia’ i. i) Ta ch´u.ng minh r˘a`ng d˜ay an do.n diˆe.u gia’ m, thˆa. t vˆa. y:
· = = = an+1 = an (n + 1)! (n + 1)n+1 n! (n + 1)n n! nn nn (n + 1)n nn (n + 1)n
v`ı
< 1 nˆen an+1 < an. nn (n + 1)n
V`ı an > 0 nˆen n´o bi. ch˘a.n du.´o.i v`a do d´o lim an tˆo` n ta. i, k´y hiˆe.u
lim an = a v`a r˜o r`ang l`a a = lim an > 0.
n
n
ii) Ta ch´u.ng minh a = 0. Thˆa. t vˆa. y ta c´o:
(cid:17) (cid:17) = 2. = (cid:16) = (cid:16)1 + > 1 + (n + 1)n nn n + 1 n 1 n n n
Do d´o:
< v`a an+1 < an. nn (n + 1)n 1 2 1 2
.c a 6 ⇒ a = 0. N Chuyˆe’n qua gi´o.i ha. n ta du.o. a 2
B `AI T ˆA. P
23 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
1. Cho c´ac d˜ay sˆo´:
· 1) an = sin n. 3) cn = n cos πn. 5n2 n2 + 3 2) bn = (−1)n 2n n + 1
H˜ay chı’ ra d˜ay n`ao bi. ch˘a. n v`a d˜ay n`ao khˆong bi. ch˘a. n. (DS. 1) v`a 2) bi. ch˘a. n; 3) khˆong bi. ch˘a.n)
2. Ch´u.ng minh r˘a`ng d˜ay:
, . . . , a1 = , a2 = , a3 = a1 a + a1 a2 a + a2
, . . . an = (a > 1, a0 > 0) a0 a + a0 an−1 a + an−1
hˆo. i tu. . 3. Ch´u.ng minh c´ac d˜ay sau dˆay hˆo. i tu.
1) an = n2 − 1 n2
+ 1 3! 1 n! .c suy t`u. n! > 2n−1 v`a do d´o 1 + · · · + 2) an = 2 + 2! Chı’ dˆa˜n. T´ınh bi. ch˘a. n du.o.
+ an 6 2 + 1 2 1 22 + · · · + 1 2n−1 = 3 − 1 2n−1 < 3.
√ (DS. k−1 5) 4. Ch´u.ng minh c´ac d˜ay sau dˆay hˆo. i tu. v`a t`ım gi´o.i ha. n a cu’ a ch´ung 5an, k ∈ N. √ 1) a1 = k
2) an =
< 1. = (DS. a = 0) Chı’ dˆa˜n. 2 n + 3
trong d´o E(nx) l`a phˆa` n nguyˆen cu’ a nx. √ 5, an+1 = k 2n (n + 2)! an+1 an E(nx) n
3) an = Chı’ dˆa˜n. Su.’ du. ng hˆe. th´u.c: nx − 1 < E(nx) 6 nx.
(DS. a = x) 5. Ch´u.ng minh r˘a`ng d˜ay: an = a1/2n hˆo. i tu. v`a t`ım gi´o.i ha. n cu’ a n´o (a > 1).
24 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
(DS. a = 1. Chı’ dˆa˜n. Ch´u.ng minh r˘a`ng an l`a d˜ay do.n diˆe.u gia’ m v`ı
√ = a1/(2n·2) = an+1 = a1/2n+1 an, an > 1)
6. Ch´u.ng minh r˘a`ng d˜ay
+ + · · · + an = 1 + 1 22 1 32 1 n2
hˆo. i tu. .
du.o. Chı’ dˆa˜n. Ch´u.ng to’ r˘a`ng d˜ay do.n diˆe.u t˘ang, t´ınh bi. ch˘a. n cu’ a n´o .c x´ac lˆa. p b˘a`ng c´ach su.’ du. ng c´ac bˆa´t d˘a’ ng th´u.c:
− = , n > 2. < 1 n2 1 n(n − 1) 1 n − 1 1 n
7. Ch´u.ng minh r˘a`ng d˜ay
+ + · · · + an = 1 3 + 1 1 32 + 2 1 3n + n
c´o gi´o.i ha. n h˜u.u ha. n.
n+1(cid:17) do.n diˆe.u gia’ m v`a (cid:1)
.c x´ac lˆa. p b˘a`ng c´ach so s´anh an Chı’ dˆa˜n. T´ınh bi. ch˘a. n cu’ a an du.o.
n+1
v´o.i tˆo’ng mˆo. t cˆa´p sˆo´ nhˆan n`ao d´o. 1 8. Ch´u.ng minh r˘a`ng d˜ay (cid:16)(cid:0)1 + n
(cid:17) (cid:16)1 + = e. lim n→∞ 1 n
n (cid:17)
an, nˆe´u 9. T´ınh lim n→∞
, k ∈ N. (DS. e) 1) an = (cid:16)1 +
n
1 n + k n (cid:17) ) . (DS. 2) an = (cid:16)
2n
1 e √ (cid:17) . (DS. e) 3) an = (cid:16)1 +
(cid:17) . (DS. e) 4) an = (cid:16) n n + 1 1 2n 2n + 1 2n
. hˆo. i tu. cu’ a d˜ay sˆo´ du.
7.1.4 Ch´u.ng minh su.
25 7.1. Gi´o.i ha. n cu’ a d˜ay sˆo´
.a trˆen diˆe`u kiˆe. n cˆa` n v`a du’ dˆe’ d˜ay hˆo. i tu. (nguyˆen l´y hˆo. i tu. Bolzano-Cauchy) . hˆo. i tu. cu’ a d˜ay. Trˆen dˆay ta d˜a nˆeu hai phu.o.ng ph´ap ch´u.ng minh su. .c dˆo´i v´o.i c´ac d˜ay khˆong do.n Hai phu.o.ng ph´ap n`ay khˆong ´ap du. ng du.o. diˆe. u du.o. .c cho b˘a`ng .c cho khˆong b˘a`ng phu.o.ng ph´ap gia’ i t´ıch m`a du.o. phu.o.ng ph´ap kh´ac (ch˘a’ ng ha. n b˘a`ng phu.o.ng ph´ap truy hˆo` i). M˘a. t . hˆo. i tu. kh´ac, trong nhiˆe`u tru.`o.ng ho. hay phˆan k`y cu’ a d˜ay m`a thˆoi. Sau dˆay ta ph´at biˆe’u mˆo. t tiˆeu chuˆa’n .a . hˆo. i tu. cu’ a d˜ay chı’ du. c´o t´ınh chˆa´t “nˆo. i ta. i” cho ph´ep kˆe´t luˆa. n su. trˆen gi´a tri. cu’ a c´ac sˆo´ ha. ng cu’ a d˜ay:
.p ngu.`o.i ta chı’ quan tˆam dˆe´n su.
Nguyˆen l´y hˆo. i tu. . D˜ay (an) c´o gi´o.i ha. n h˜u.u ha. n khi v`a chı’ khi n´o tho’a m˜an diˆe`u kiˆe.n:
∀ ε > 0, ∃ N0 = N0(ε) ∈ N : ∀ n > N0 v`a ∀ p ∈ N ⇒ |an − an+p| < ε.
T`u. nguyˆen l´y hˆo. i tu. r´ut ra: D˜ay (an) khˆong c´o gi´o.i ha. n khi v`a chı’
khi n´o tho’a m˜an diˆe`u kiˆe. n:
∃ ε > 0, ∀ N ∈ N ∃ n > N ∃ m > N → |an − am| > ε.
C ´AC V´I DU.
V´ı du. 1. Ch´u.ng minh r˘a`ng d˜ay
+ + · · · + , n ∈ N an = cos 1 3 cos 2 32 cos n 3n
hˆo. i tu. .
26 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Gia’ i. Ta u.´o.c lu.o. .ng hiˆe.u
+ · · · + cos(n + 1) 3n+1 cos(n + p) 3n+p |an+p − an| = (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
+ · · · + 6 1 3n+p 1 3n+1
1 − · · = < < 1 3n+1 1 2 1 3n 1 3n 1 − 1 3p 1 3
1 Gia’ su.’ ε l`a sˆo´ du.o.ng t`uy ´y. V`ı
3n = 0 nˆen v´o.i sˆo´ ε > 0 d´o, 1 3n < ε. Ngh˜ıa l`a nˆe´u n > N , . nhiˆen t`uy ´y th`ı lim n→∞ tˆo` n ta. i sˆo´ N ∈ N sao cho ∀ n > N ta c´o c`on p l`a sˆo´ tu.
< ε. |an+p − an| < 1 3n
Do d´o theo tiˆeu chuˆa’n hˆo. i tu. d˜ay d˜a cho hˆo. i tu. . N V´ı du. 2. Ch´u.ng minh r˘a`ng d˜ay
+ + · · · + an = 1 √ n 1 √ 1 1 √ 2
phˆan k`y.
Gia’ i. Ta u.´o.c lu.o.
√ √ √ + + · · · + 1 n + p 1 n + 2 (cid:12) (cid:12) (cid:12)
|an − an+p| = (cid:12) (cid:12) (cid:12) √ > ∀ n, p ∈ N. .ng hiˆe.u 1 n + 1 p n + p
D˘a. c biˆe.t v´o.i p = n ta c´o
√ √ ∀ n. (*) > |an − a2n| > n 2 1 √ 2
Ta lˆa´y ε = 1 √ 2
. Khi d´o ∀ N ∈ N tˆo` n ta. i nh˜u.ng gi´a tri. n > N v`a ∃ p ∈ N sao cho |an − an+p| > ε. Thˆa. t vˆa. y, theo bˆa´t d˘a’ ng th´u.c (*) ta
27 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
chı’ cˆa` n lˆa´y sˆo´ n > N bˆa´t k`y v`a p = n. T`u. d´o theo mˆe.nh dˆe` phu’ di.nh nguyˆen l´y hˆo. i tu. ta c´o d˜ay d˜a cho phˆan k`y. N
B `AI T ˆA. P
. hˆo. i tu. cu’ a d˜ay (an) Su.’ du. ng tiˆeu chuˆa’n hˆo. i tu. dˆe’ ch´u.ng minh su. nˆe´u
n P k=1
, α ∈ R. 1. an = sin nα 2n
n P k=1
2. an = akqk, |q| < 1, |ak| < M ∀ k, M > 0.
n P k=1
· 3. an = (−1)k−1 k(k + 1)
n P k=1
· 4. an = (−1)k k!
. 5. an = 0, 77 . . . 7 | {z } nch˜u. sˆo´
n P k=1
· 6. an = 1 2k + k
, n ∈ N. + · · · + 7. an = 1 + Ch´u.ng minh r˘a`ng c´ac d˜ay sau dˆay phˆan k`y: 1 n 1 2
+ + · · · + , n = 2, . . . 8. an = 1 ln2 1 ln3 1 lnn
.c ph´at biˆe’u nhu. sau.
7.2 Gi´o.i ha. n h`am mˆo. t biˆe´n 7.2.1 C´ac kh´ai niˆe. m v`a di.nh l´y co. ba’ n vˆe` gi´o.i ha. n Di.nh ngh˜ıa gi´o.i ha. n cu’ a c´ac h`am dˆo´i v´o.i n˘am tru.`o.ng ho. .p: x → a, x → a ± 0, x → ±∞ du.o.
28 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
1) Sˆo´ A du.o. .c go. i l`a gi´o.i ha. n cu’ a h`am f (x) ta. i diˆe’m a (khi x → a) .c sˆo´ δ = δ(ε) > 0 (∃δ = δ(ε) >
nˆe´u ∀ ε > 0 b´e bao nhiˆeu t`uy ´y t`ım du.o. 0) sao cho ∀ x m`a
x ∈ Df ∩ {x; 0 < |x − a| < δ(ε)}
th`ı
|f (x) − A| < ε.
f (x) = A.
lim K´y hiˆe.u: x→a .c go. i l`a gi´o.i ha. n bˆen pha’ i (bˆen tr´ai) cu’ a h`am f (x) ta. i 2) Sˆo´ A du.o. diˆe’m x = a nˆe´u ∀ ε > 0, ∃ δ = δ(ε) > 0 sao cho v´o.i mo. i x tho’a m˜an diˆe`u kiˆe.n
x ∈ Df ∩ {x : a < x < a + δ} (x ∈ Df ∩ {x : a − δ < x < a})
th`ı
|f (x) − A| < ε.
K´y hiˆe.u:
f (x) = f (a − 0)(cid:17). f (x) = f (a + 0) (cid:16) lim x→a−0
lim x→a+0 .: Tu.o.ng tu. f (x) = A ⇔ ∀ ε > 0 ∃ ∆ > 0 : ∀ x ∈ Df ∩ {x : x > ∆} lim 3) x→+∞
⇒ |f (x) − A| < ε.
.. .c ph´at biˆe’u tu.o.ng tu. f (x) = A th`ı ngu.`o.i ta viˆe´t Di.nh ngh˜ıa gi´o.i ha. n khi x → −∞ du.o. 4) Nˆe´u lim f (x) = lim x→−∞ x→+∞
f (x) = A. lim x→∞
29 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
Tru.`o.ng ho. .c go. i l`a h`am vˆo .p d˘a. c biˆe.t nˆe´u A = 0 th`ı h`am f (x) du.o. c`ung b´e khi x → a (x → a ± 0, x → ±∞).
.c ph´at biˆe’u dˆo´i Kh´ai niˆe.m h`am vˆo c`ung l´o.n ta. i diˆe’m a c˜ung du.o. .p. v´o.i ca’ n˘am tru.`o.ng ho.
.c go. i l`a h`am vˆo c`ung l´o.n ta. i diˆe’m a nˆe´u Ch˘a’ ng ha. n, h`am f (x) du.o.
∀ M > 0 ∃ δ = δ(M) > 0 : ∀ x ∈ Df ∩ {x : 0 < |x − a| < δ} ⇒ |f (x)| > M.
Ngo`ai ra, nˆe´u f (x) > 0 (f (x) < 0) ∀ x ∈ Df ∩ {x : 0 < |x − a| < δ}
th`ı ta viˆe´t
f (x) = −∞(cid:17). lim x→a f (x) = +∞ (cid:16) lim x→a
Ta lu.u ´y r˘a`ng c´ac k´y hiˆe.u v`u.a nˆeu chı’ ch´u.ng to’ f (x) l`a vˆo c`ung
l´o.n ch´u. ho`an to`an khˆong c´o ngh˜ıa r˘a`ng f c´o gi´o.i ha. n.
x→a
Khi t´ınh gi´o.i ha. n ta thu.`o.ng su.’ du. ng c´ac diˆe`u kh˘a’ ng di.nh sau dˆay. f2(x) tˆo` n ta. i h˜u.u ha. n f1(x), lim x→a D- i.nh l´y 7.2.1. Nˆe´u c´ac gi´o.i ha. n lim th`ı
f2(x) f1(x) + lim x→a
f2(x) 1) lim x→a 2) lim x→a [f1(x) + f2(x)] = lim x→a [f1(x) · f2(x)] = lim x→a f1(x) · lim x→a
f1(x) = 3) Nˆe´u lim x→a f2(x) 6= 0 th`ı lim x→a f1(x) f2(x) f2(x) lim x→a lim x→a
4) Nˆe´u trong lˆan cˆa. n U (a; δ) = {x : 0 < |x − a| < δ} ta c´o f (x) = A f1(x) = lim x→a f2(x) = A th`ı lim x→a f1(x) 6 f (x) 6 f2(x) v`a lim x→a (nguyˆen l´y bi. ch˘a. n hai phi´a).
Di.nh ngh˜ıa gi´o.i ha. n h`am sˆo´ c´o thˆe’ ph´at biˆe’u du.´o.i da. ng ngˆon ng˜u.
d˜ay nhu. sau. D- i.nh l´y 7.2.2. Gia’ su.’ D ⊂ R, a ∈ R l`a diˆe’m tu. cu’ a n´o; A ∈ R, f : D → R. Khi d´o
f (x) = A lim x→a
30 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
khi v`a chı’ khi ∀(an), an ∈ D \ {a}, an → a
f (an) → A
T`u. d´o dˆe’ ch´u.ng minh mˆo. t h`am n`ao d´o khˆong c´o gi´o.i ha. n khi x → a,
n) dˆe`u hˆo. i tu. dˆe´n a nhu.ng
ta chı’ cˆa` n ch´u.ng minh r˘a`ng ∃(an), ∃(a0
f 0(an). lim x→a f (an) 6= lim x→a
C´ac di.nh l´y co. ba’ n vˆe` gi´o.i ha. n d˜a ph´at biˆe’u trˆen dˆay khˆong ´ap
du. ng du.o.
.c dˆo´i v´o.i c´ac gi´o.i ha. n sau dˆay khi x → a, a ∈ R. [f (x)+g(x)]; f , g l`a c´ac vˆo c`ung l´o.n (vˆo di.nh da. ng “∞±∞”). 1) lim x→a
; f , g ho˘a. c dˆo` ng th`o.i l`a hai vˆo c`ung b´e, ho˘a. c dˆo` ng th`o.i 2) lim x→a f (x) g(x)
l`a hai vˆo c`ung l´o.n (vˆo di.nh da. ng “0/0” ho˘a. c “∞/∞”).
g(x)
.c f (x) · g(x); f l`a vˆo c`ung b´e, c`on g l`a vˆo c`ung l´o.n ho˘a. c ngu.o. 3) lim x→a la. i (vˆo di.nh da. ng “0 · ∞”).
: (cid:2)f (x)(cid:3)
.p n`ay thu.`o.ng du.o. .c go. i .p khi t´ınh gi´o.i ha. n ta
4) lim x→a a) khi f (x) → 1, g(x) → ∞ (vˆo di.nh da. ng “1∞”) b) khi f (x) → 0, g(x) → 0 (vˆo di.nh da. ng “00”) c) khi f (x) → ∞, g(x) → 0 (vˆo di.nh da. ng “∞0”) Viˆe.c t´ınh gi´o.i ha. n trong c´ac tru.`o.ng ho. l`a khu.’ da. ng vˆo di.nh. Trong nhiˆe`u tru.`o.ng ho. thu.`o.ng su.’ du. ng c´ac gi´o.i ha. n quan tro. ng sau dˆay:
1
= 1, (7.12) lim x→0
x = e
sin x x (1 + x) (7.13) lim x→0
31 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
x
v`a c´ac hˆe. qua’ cu’ a (7.13)
(cid:17) = e, (7.14) (cid:16)1 + lim x→∞ 1 x
= , 0 < a 6= 1, (7.15) lim x→0 1 lna
= lna, 0 < a 6= 1. (7.16) lim x→0 loga(1 + x) x ax − 1 x
C ´AC V´I DU. V´ı du. 1. Su.’ du. ng (ε − δ) - di.nh ngh˜ıa gi´o.i ha. n dˆe’ ch´u.ng minh r˘a`ng
x2 = 9. lim x→−3
Gia’ i. Ta cˆa` n ch´u.ng minh r˘a`ng ∀ ε > 0, ∃ δ > 0 sao cho v´o.i
|x + 3| < δ th`ı ta c´o |x2 − 9| < ε.
Ta cˆa` n u.´o.c lu.o. .ng hiˆe.u |x2 − 9|. ta c´o
|x2 − 9| = |x − 3||x + 3|.
Do th`u.a sˆo´ |x − 3| khˆong bi. ch˘a. n trˆen to`an tru. c sˆo´ nˆen dˆe’ u.´o.c lu.o. .ng t´ıch do.n gia’ n ho.n ta tr´ıch ra 1 - lˆan cˆa. n cu’ a diˆe’m a = −3 t´u.c l`a khoa’ ng (−4; −2). V´o.i mo. i x ∈ (−4; −2) ta c´o |x − 3| < 7 v`a do d´o
|x2 − 9| < 7|x + 3|.
x→−3
V`ı δ-lˆan cˆa. n diˆe’m a = −3 [t´u.c l`a khoa’ ng (−3 − δ; −3 + δ)] khˆong ε .t ra kho’i ranh gi´o.i cu’ a 1-lˆan cˆa. n nˆen ta lˆa´y δ = min (cid:16)1, .c vu.o. (cid:17). 7 x2 = 9. N
x→2
√ 11 − x = 3. du.o. Khi d´o v´o.i 0 < |x + 3| < δ ⇒ |x2 − 9| < ε. Do vˆa. y lim V´ı du. 2. Ch´u.ng minh r˘a`ng lim
Gia’ i. Gia’ su.’ ε > 0 l`a sˆo´ du.o.ng cho tru.´o.c b´e bao nhiˆeu t`uy ´y. Ta
x´et bˆa´t phu.o.ng tr`ınh
√ | 11 − x − 3| < ε. (7.17)
32 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Ta c´o
√ √ 11 − x − 3 > −ε (7.17) ⇔ −ε < 11 − x − 3 < ε ⇔ √ 11 − x − 3 < ε
x − 11 < −(3 − ε)2 x − 2 < 6ε − ε3 ⇔ ⇔ x − 11 > −(3 + ε)2 x − 2 > −(6ε + ε2).
√ V`ı 6ε − ε2 < | − (6ε + ε)2| = 6ε + ε2 nˆen ta c´o thˆe’ lˆa´y δ(ε) l`a sˆo´ δ 6 6ε − ε2. V´o.i sˆo´ δ d´o ta thˆa´y r˘a`ng khi x tho’a m˜an bˆa´t d˘a’ ng th´u.c 0 < |x − 2| < δ th`ı | 11 − x − 3| < ε v`a √ 11 − x = 3. N
); 1) lim x→2 0 0
x (cid:17)
lim x→2 V´ı du. 3. T´ınh c´ac gi´o.i ha. n 2x − x2 x − 2 cotg2x · cotg(cid:16) 2) − x(cid:17) (vˆo di.nh da. ng 0 · ∞); (vˆo di.nh da. ng π 4 lim x→ π 4
x +
(cid:16)e 1 3) (vˆo di.nh da. ng 1∞). lim x→∞ 1 x
Gia’ i 1) Ta c´o
− · = = 4 · 2x − x2 x − 2 2x − 22 − (x2 − 22) x − 2 2x−2 − 1 x − 2 x2 − 4 x − 2
T`u. d´o suy r˘a`ng
= 4ln2 − 4. lim x→2 = 4 lim x→2 − lim x→2 2x−2 − 1 x − 2 x2 − 4 x − 2
− x. Khi d´o 2) D˘a. t y = 2x − x2 x − 2 π 4
− 2y(cid:17)cotgy cotg2x · cotg(cid:16) cotg(cid:16) − x(cid:17) = lim y→0 π 4 π 2 lim x→ π 4
· = 2. = lim y→0 sin 2y sin y cos y cos 2y
33 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
x
1
1
ln(ey +y) y
lim y→0
. Khi d´o 3) D˘a. t y = 1 x
x +
y = e
(cid:17) (ey + y) ; (cid:16)e lim x→∞ = lim y→0 1 x
· lim y→0 = lim y→0 ln(ey + y) y ln[1 + (ey + y − 1)] ey + y − 1
(cid:16)1 + (cid:17) = 2. = lim t→0 · lim y→0 ln(1 + t) t ey + y − 1 y ey − 1 y
1
y = e2. N
T`u. d´o suy r˘a`ng
(cid:0)ey + y(cid:1) lim y→0
khˆong c´o gi´o.i ha. n khi 1 x V´ı du. 4. Ch´u.ng to’ r˘a`ng h`am f (x) = sin x → 0.
Gia’ i. Ta lu.u ´y mˆe.nh dˆe` phu’ di.nh dˆo´i v´o.i di.nh ngh˜ıa gi´o.i ha. n:
f (x) 6= A ⇔ ∃ ε0 > 0 ∀ δ > 0 ∃ xδ (0 < |xδ − a| < δ) lim x→a
→ |f (x0) − A| > ε0.
2 . Khi d´o ∀ δ > 0, Nˆe´u A = 0 ta lˆa´y ε0 = v`a xk = 1 2 + 2kπ π 2 ∃ k ∈ N : 0 < xk < δ v`a
Nˆe´u A 6= 0 th`ı ta lˆa´y ε0 = v`a xk = 1 2kπ
khi x → 0. N 1 x
|f (xk) − 0| = |f (xk)| = 1 > ε0 v`a nhu. vˆa. y A = 0 khˆong pha’ i l`a gi´o.i ha. n cu’ a h`am d˜a cho khi x → 0. |A| . Khi d´o ∀ δ > 0, 2 ∃ k ∈ N : 0 < xk < δ th`ı |f (xk) − A| = |A| > ε. Nhu. vˆa. y mo. i sˆo´ A 6= 0 dˆe`u khˆong l`a gi´o.i ha. n cu’ a h`am sin V´ı du. 5. H`am Dirichlet D(x):
1 nˆe´u x ∈ Q, D(x) = 0 nˆe´u x ∈ R \ Q
34 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
khˆong c´o gi´o.i ha. n ta. i ∀ a ∈ R.
n).
D(a0 Gia’ i. Ta ch´u.ng minh r˘a`ng ta. i mo. i diˆe’m a ∈ R h`am D(x) khˆong tho’a m˜an Di.nh l´y 2. Dˆe’ l`am viˆe.c d´o, ta chı’ cˆa` n chı’ ra hai d˜ay (an) v`a (a0 D(an) 6= lim n→∞
n) c`ung hˆo. i tu. dˆe´n a sao cho lim n→∞ Dˆa` u tiˆen ta x´et d˜ay c´ac diˆe’m h˜u.u ty’ (an) hˆo. i tu. dˆe´n a. Ta c´o n) - n) = 0 ∀ n v`a do vˆa. y
D(an) = 1. Bˆay gi`o. ta x´et d˜ay (a0
n). T`u. d´o suy ra r˘a`ng ta. i diˆe’m a
n→∞
D(an) = 1 ∀ n v`a do d´o lim n→∞ d˜ay c´ac diˆe’m vˆo ty’ hˆo. i tu. dˆe´n a. Ta c´o D(a0 lim n→∞ D(a0 D(a0 n) = 0. Nhu. vˆa. y lim D(an) 6= lim n→∞
g(x) = +∞. Ch´u.ng minh r˘a`ng h`am D(x) khˆong c´o gi´o.i ha. n . N V´ı du. 6. Gia’ su.’ f (x) = b, lim x→a lim x→a
[f (x) + g(x)] = +∞. lim x→a
Gia’ i. Ta cˆa` n ch´u.ng minh r˘a`ng ∀ M > 0, ∃ δ > 0 sao cho ∀ x : 0 <
|x − a| < δ th`ı f (x) + g(x) > M.
f (x) = b nˆen tˆo` n ta. i δ1-lˆan cˆa. n U (a, δ1) cu’ a diˆe’m a sao cho V`ı lim x→a
|f (x)| < C, x 6= a (7.18)
trong d´o C l`a h˘a`ng sˆo´ du.o.ng n`ao d´o.
g(x) = +∞ nˆen dˆo´i Gia’ su.’ M > 0 l`a sˆo´ cho tru.´o.c t`uy ´y. V`ı lim x→a v´o.i sˆo´ M + C, ∃ δ > 0 (δ 6 δ1) sao cho ∀ x : 0 < |x − a| < δ th`ı
g(x) > M + C (7.19)
.c l`a: v´o.i x tho’a T`u. c´ac bˆa´t d˘a’ ng th´u.c (7.18) v`a(7.19) ta thu du.o.
m˜an diˆe`u kiˆe.n 0 < |x − a| < δ 6 δ1 th`ı
f (x) + g(x) > g(x) − |f (x)| > M + C − C = M. N
B `AI T ˆA. P
35 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
1. Su.’ du. ng di.nh ngh˜ıa gi´o.i ha. n h`am sˆo´ dˆe’ ch´u.ng minh c´ac d˘a’ ng th´u.c sau dˆay:
sin x = sin x = 1; 1 2 1) lim x→ π 6 ; 2) lim x→ π 2
. = 0; 4) arctgx = x sin 3) lim x→0 π 2
) − arctgx < tg(cid:0) − arctgx(cid:1) = lim x→+∞ π 2 π 2 1 x
; 6) = logax = +∞; 5) lim x→∞ lim x→+∞ 1 3
7) = −8; 1 x Chı’ dˆa˜n. D`ung hˆe. th´u.c x − 1 3x + 2 √ (cid:0) lim x→+∞ x2 + 1 − x(cid:1) = 0; 8) lim x→−5
= ; 9) lim x→1 (5x2 − 7x + 6) = 4; 10) lim x→2 x2 + 2x − 15 x + 5 x2 − 3x + 2 x2 + x − 6 1 5
11) = 0. lim x→+∞ x sin x x2 − 100x + 3000
; sin x; 2 1 x ; sin 2. Ch´u.ng minh c´ac gi´o.i ha. n sau dˆay khˆong tˆo` n ta. i: 2) lim x→∞ 1) lim x→1 3) lim x→o 1 x − 1
e 1 x ; cos x. 5) lim x→∞
4) lim x→0 Nˆe´u tu.’ sˆo´ v`a mˆa˜u sˆo´ cu’ a phˆan th´u.c h˜u.u ty’ dˆe`u triˆe.t tiˆeu ta. i diˆe’m x = a th`ı c´o thˆe’ gia’ n u.´o.c phˆan th´u.c cho x − a (6= 0) mˆo. t ho˘a. c mˆo. t sˆo´ lˆa` n.
Su.’ du. ng phu.o.ng ph´ap gia’ n u.´o.c d´o, h˜ay t´ınh c´ac gi´o.i ha. n sau dˆay (3-10).
3. ) (DS. lim x→7 17 5
4. (DS. 2) lim x→1 2x2 − 11x − 21 x2 − 9x + 14 x4 − x3 + x2 − 3x + 2 x3 − x2 − x + 1
5. (DS. −8) lim x→1
; m, n ∈ Z ) 6. (DS. lim x→1 x4 + 2x2 − 3 x2 − 3x + 2 xm − 1 xn − 1 m n
36 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
(cid:16) (cid:17) − 7. (DS. −1) lim x→1
(cid:16) − (cid:17); a, b ∈ N ) (DS. 8. lim x→1 1 1 − x a 1 − xa 3 1 − x3 b 1 − xb a − b 2
9. (DS. C k n) lim x→1 (xn − 1)(xn−1 − 1) · · · (xn−k+1 − 1) (x − 1)(x2 − 1) · · · (xk − 1)
, n ∈ N (DS. an−1) 10. lim x→a (xn − an) − nan−1(x − a) (x − a)2 n(n − 1) 2
p
Chı’ dˆa˜n. Dˆo’i biˆe´n x − a = t. C´ac b`ai to´an sau dˆay c´o thˆe’ du.a vˆe` da. ng trˆen nh`o. ph´ep dˆo’i biˆe´n (11-14)
) (DS. 11. lim x→1 ps qr
q − 1 x x r s − 1 √ 1 + 3 √ 1 + 5 √ 3 3
) (DS. 12. lim x→−1 5 3 x x
n
1 + x + 1 ) 13. (DS. lim x→0 1 6 √ 1 + x − 4 4 √ 2 − 2 1 + x + x √
14. ) (DS. lim x→0 1 + x − 1 x 1 n
Mˆo. t trong c´ac phu.o.ng ph´ap t´ınh gi´o.i ha. n cu’ a c´ac biˆe’u th´u.c vˆo ty’
l`a chuyˆe’n vˆo ty’ t`u. mˆa˜u sˆo´ lˆen tu.’ sˆo´ ho˘a. c ngu.o. .c la. i (15-26) √
) (DS. 15. lim x→0 √ √ 1 + x + x2 − 1 x 3 + x + x2 − 1 2 9 − 2x + x2 ) (DS. 16. lim x→2 1 2
3
3
x2 − 3x + 2 5x √ 17. ) (DS. lim x→0 15 2 √ 1 + x − 3 √ 1 − 2x 18. (DS. 2) lim x→0 1 − x √ 1 + 3x − 3 x + x2 √ x2 + 1 − (DS. 0) 19. √ (cid:0) x2 − 1(cid:1) lim x→∞
37 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
(DS. 0) 20. √ (cid:0) 3 lim x→∞
21. (DS. +∞) x2 + 5x + x(cid:1) 1 − x3 + x(cid:1) √ (cid:0) lim x→+∞
) (DS. − 22. x2 + 5x + x(cid:1) lim x→−∞ 5 2
23. (DS. 1) x2 + 2x − x(cid:1) lim x→+∞
3 i
24. (DS. +∞) √ (cid:0) √ (cid:0) √ (cid:0) x2 + 2x − x(cid:1). lim x→−∞
3 − (x − 1) 2
h(x + 1) 2 25. (DS. 0) lim x→∞
26. (cid:2) np(x + a1)(x + a2) · · · (x + an) − x(cid:3) lim x→+∞
(DS. a1 + a2 + · · · + an n
) Khi gia’ i c´ac b`ai to´an sau dˆay ta thu.`o.ng su.’ du. ng hˆe. th´u.c
5
= α (27-34) (1 + t)α − 1 t √
3
n
n
−1)
4
3
lim t→0 √ √ (DS. −6) 27. lim x→0 √ 1 + 3x4 − √ 1 + x − √ a + x − n 1 − 2x 1 + x a − x 28. a 1 lim x→0 x √ √ 1 + 3x + 3 2 n 1 + x √ ) (DS. 29. lim x→0 313 280 √
n 1 + x2 − x(cid:1)
n
n − (cid:0) x a − x
n
, n ∈ N (DS. √ √ 1 + x − 5 1 + x − 7 √ 1 + 2x + x − 6 1 + x √ a2 + ax + x2 − 3 √ √ (DS. a 1 6 ) 30. lim x→0 3 2 a + x − a2 − ax + x2 a − x √ 1 + x2 + x(cid:1) √ (cid:0) (DS. 2n) 31. lim x→0 √ √ a + x − n , n ∈ N, a > 0 ) (DS. 32. lim x→0 x √ √ 1 + ax − k 1 + bx 33. , n ∈ N, a > 0 ) (DS. lim x→0 x
) (DS. 34. (cid:0) np(1 + x2)(2 + x2) · · · (n + x2) − x2(cid:1) lim x→∞ √ 2 n a na ak − bn nk n + 1 2
38 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Khi t´ınh gi´o.i ha. n c´ac biˆe’u th´u.c lu.o. .ng gi´ac ta thu.`o.ng su.’ du. ng cˆong th´u.c co. ba’ n
= 1 lim x→0 sin x x
. kˆe´t ho. .p c´ac phu.o.ng ph´ap t`ım gi´o.i ha. n d˜a nˆeu o.’ trˆen c`ung v´o.i su. (35-56).
sin πx 2 (DS. 0) 35. lim x→∞
36. (DS. 0) lim x→∞ x arctgx 2x
37. (DS. −4) lim x→−2 x2 − 4 arctg(x + 2)
) (DS. 38. lim x→0 tgx − sin x x3 1 2
39. ) xcotg5x (DS. lim x→0 1 5
) (1 − x)tg (DS. 40. lim x→1 πx 2 2 π
41. ) (DS. lim x→1 2 π
) (DS. 42. lim x→π 1 2π
43. (DS. (n2 − m2)) lim x→0 1 2
i − cos (DS. 4) x2h cos 44. lim x→∞ 1 − x2 sin πx sin x π2 − x2 cos mx − cos nx x2 1 x
45. (DS. − sin a) lim x→0
(DS. −2 cos a) 46. lim x→0 √ 3 x sin(a + x) + sin(a − x) − 2 sin a x2 cos(a + x) + cos(a − x) − 2 cos a 1 − cos x √ 47. x2 + 1 − sin (DS. 0) (cid:0) sin x2 − 1(cid:1) lim x→∞
39 7.2. Gi´o.i ha. n h`am mˆo. t biˆe´n
√
) 48. (DS. − lim x→0 1 4
cos − sin x 2 49. ) (DS. cos x − 1 x2 x 2 cos x 1 √ 2 lim x→ π 2
(cid:17) sin (cid:16)x − 50. ) (DS. 1 √ 3 lim x→ π 3 π 3 1 − 2 cos x √
m
) 51. (DS. 2 cos x − 1 1 − tg2x 1 4 lim x→ π 4 √ √ 1 + tgx − 1 − tgx 52. (DS. 1) lim x→0 √ sin x √ cos αx − m cos βx 53. ) (DS. lim x→0 β2 − α2 2m
x2 cos x ) (DS. − 54. lim x→0 1 3
√ cos x − 3 sin2 x √ 1 − cos x cos 2x ) (DS. 55. lim x→0 tgx2 3 2 √
56. (DS. 4) lim x→0 1 + x sin x − cos x sin2 x 2
x→a
[f (x)]ϕ(x), trong d´o
Dˆe’ t´ınh gi´o.i ha. n lim f (x) → 1, ϕ(x) → ∞ khi x → a ta c´o thˆe’ biˆe´n dˆo’i biˆe’u th´u.c
ϕ(x)[f (x)−1]
1
f (x)−1 o
[f (x)]ϕ(x) nhu. sau:
ϕ(x)[f (x)−1]
x→a
n[1 + (f (x) − 1)] lim x→a
[f (x)]ϕ(x) = lim x→a = e lim
.c t´ınh theo c´ac phu.o.ng ph´ap d˜a nˆeu trˆen ϕ(x)[f (x) − 1] du.o.
ϕ(x)[f (x) − 1] = A th`ı o.’ dˆay lim x→a dˆay. Nˆe´u lim x→a
[f (x)]ϕ(x) = eA (57-68). lim x→a
x+1
40 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
x4
(cid:17) (cid:16) 57. (DS. e) lim x→∞
(cid:17) (cid:16) (DS. 0) 58. lim x→∞
59. 2x + 3 2x + 1 x2 − 1 x2 (1 + tgx)cotgx (DS. e) lim x→0
1 x2
(1 + 3tg2x)cotg2x (DS. e3) 60. lim x→0
1 cotgx
(cid:16) (cid:17) 61. (DS. e 3 2 ) lim x→0 cos x cos 2x
(sin x) (DS. −1) 62. lim x→ π 2
cotg2x
(tgx)tg2x (DS. e−1) 63. lim x→ π 2
1 x2
+ x(cid:17)i (DS. e) htg(cid:16) 64. lim x→0 π 4
1 sin2 x
65. (DS. e− 1 2 ) (cid:0) cos x(cid:1) lim x→0
1 sin x
(DS. e− 9 2 ) 66. (cid:0) cos 3x(cid:1) lim x→0
tg22x
(cid:16) (cid:17) (DS. 1) 67. lim x→0 1 + tgx 1 + sin x
68. (DS. e− 1 2 ) (cid:0) sin 2x(cid:1)
lim x→ π 4 Khi t´ınh gi´o.i ha. n c´ac biˆe’u th´u.c c´o ch´u.a h`am lˆodarit v`a h`am m˜u ta thu.`o.ng su.’ du. ng c´ac cˆong th´u.c (7.15) v`a (7.16) v`a c´ac phu.o.ng ph´ap t´ınh gi´o.i ha. n d˜a nˆeu o.’ trˆen (69-76).
69. (DS. e−1) lim x→e
) (DS. 70. lim x→10 lnx − 1 x − e lgx − 1 x − 10 1 10ln10
√ (DS. 2) 71. lim x→0 ex2 − 1 1 + sin2 x − 1
) (DS. 72. lim x→0 3 2 ex2 − cos x sin2 x
41 7.3. H`am liˆen tu. c
73. (DS. 1) lim x→0
(DS. 2) 74. lim x→0
1
x , a > 0, b > 0
, a > 0, b > 0 ln ) (DS. − 75. lim x→0 1 2 a b √ h i 76. (DS. ab) lim x→0 eαx − eβx sin αx − sin βx esin 5x − esin x ln(1 + 2x) ax2 − bx2 ln cos 2x asin x + bsin x 2
7.3 H`am liˆen tu. c D- i.nh ngh˜ıa 7.3.1. H`am f (x) x´ac di.nh trong lˆan cˆa. n cu’ a diˆe’m x0 du.o.
.c go. i l`a liˆen tu. c ta. i diˆe’m d´o nˆe´u
f (x) = f (x0). lim x→x0
Di.nh ngh˜ıa 7.3.1 tu.o.ng du.o.ng v´o.i
D- i.nh ngh˜ıa 7.3.1∗. H`am f (x) x´ac di.nh trong lˆan cˆa. n cu’ a diˆe’m x0 du.o. .c go. i l`a liˆen tu. c ta. i diˆe’m x0 nˆe´u
∀ ε > 0 ∃ δ > 0 ∀ x ∈ Df : |x − x0| < δ ⇒ |f (x) − f (x0)| < ε.
Hiˆe.u x − x0 = ∆x du.o.
.c go. i l`a sˆo´ gia cu’ a dˆo´i sˆo´, c`on hiˆe.u f (x) − .c go. i l`a sˆo´ gia cu’ a h`am sˆo´ ta. i x0 tu.o.ng ´u.ng v´o.i sˆo´ f (x0) = ∆f du.o. gia ∆x, t´u.c l`a
∆x = x − x0, ∆f (x0) = f (x0 + ∆x) − f (x0).
V´o.i ngˆon ng˜u. sˆo´ gia di.nh ngh˜ıa 7.3.1 c´o da. ng
D- i.nh ngh˜ıa 7.3.1∗∗. H`am f (x) x´ac di.nh trong lˆan cˆa. n cu’ a diˆe’m x0 du.o. .c go. i l`a liˆen tu. c ta. i x0 nˆe´u
∆f = 0. lim ∆x→0
42 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
B˘a`ng “ngˆon ng˜u. d˜ay” ta c´o di.nh ngh˜ıa tu.o.ng du.o.ng
D- i.nh ngh˜ıa 7.3.1∗∗∗. H`am f (x) x´ac di.nh trong lˆan cˆa. n diˆe’m x0 ∈ Df du.o. .c go. i l`a liˆen tu. c ta. i diˆe’m x0 nˆe´u
f (xn) = f (x0). ∀(xn) ∈ Df : xn → x0 ⇒ lim n→∞
D- i.nh l´y 7.3.1. Diˆe`u kiˆe. n cˆa` n v`a du’ dˆe’ h`am f (x) liˆen tu. c ta. i diˆe’m x0 l`a h`am f (x) tho’a m˜ac c´ac diˆe`u kiˆe. n sau dˆay:
i) H`am pha’ i x´ac di.nh ta. i mˆo. t lˆan cˆa. n n`ao d´o cu’ a diˆe’m x0. ii) H`am c´o c´ac gi´o.i ha. n mˆo. t ph´ıa nhu. nhau
x→x0+0
f (x) = lim f (x). lim x→x0−0
x→x0+0
= lim = f (x0). lim x→x0−0
iii) Gia’ su.’ h`am f (x) x´ac di.nh trong nu.’ a lˆan cˆa. n bˆen pha’ i (bˆen tr´ai) cu’ a diˆe’m x0, ngh˜ıa l`a trˆen nu.’ a khoa’ ng [x0, x0 + δ) (tu.o.ng ´u.ng: trˆen (x0 − δ, x0]) n`ao d´o. H`am f (x) du.o. .c go. i l`a liˆen tu. c bˆen pha’ i (bˆen tr´ai) ta. i diˆe’m x0 nˆe´u
f (x0 + 0) = f (x0) (tu.o.ng ´u.ng: f (x0 − 0) = f (x0)). D- i.nh l´y 7.3.2. H`am f (x) liˆen tu. c ta. i diˆe’m x0 ∈ Df khi v`a chı’ khi n´o liˆen tu. c bˆen pha’ i v`a bˆen tr´ai ta. i diˆe’m x0.
H`am liˆen tu. c ta. i mˆo. t diˆe’m c´o c´ac t´ınh chˆa´t sau. I) Nˆe´u c´ac h`am f (x) v`a g(x) liˆen tu. c ta. i diˆe’m x0 th`ı f (x) ± g(x), f (x) · g(x) liˆen tu. c ta. i x0, v`a f (x)/g(x) liˆen tu. c ta. i x0 nˆe´u g(x0) 6= 0. II) Gia’ su.’ h`am y = ϕ(x) liˆen tu. c ta. i x0, c`on h`am u = f (y) liˆen
.p u = f [ϕ(x)] liˆen tu. c ta. i x0. tu. c ta. i y0 = ϕ(x0). Khi d´o h`am ho. T`u. d´o suy ra r˘a`ng
ϕ(x)(cid:3). lim x→x0 f [ϕ(x)] = f (cid:2) lim x→x0
H`am f (x) go. i l`a gi´an doa. n ta. i diˆe’m x0 nˆe´u n´o x´ac di.nh ta. i nh˜u.ng diˆe’m gˆa` n x0 bao nhiˆeu t`uy ´y nhu.ng ta. i ch´ınh x0 h`am khˆong tho’a m˜an ´ıt nhˆa´t mˆo. t trong c´ac diˆe`u kiˆe.n liˆen tu. c o.’ trˆen.
43 7.3. H`am liˆen tu. c
.c go. i l`a
Diˆe’m x0 du.o. 1) Diˆe’m gi´an doa. n khu.’ du.o. .c cu’ a h`am f (x) nˆe´u tˆo` n ta. i lim x→x0
.c. f (x) = b nhu.ng ho˘a. c f (x) khˆong x´ac di.nh ta. i diˆe’m x0 ho˘a. c f (x0) 6= b. Nˆe´u bˆo’ sung gi´a tri. f (x0) = b th`ı h`am f (x) tro.’ nˆen liˆen tu. c ta. i x0, t´u.c l`a gi´an doa. n c´o thˆe’ khu.’ du.o.
2) Diˆe’m gi´an doa. n kiˆe’u I cu’ a h`am f (x) nˆe´u ∃ f (x0+0) v`a ∃ f (x0−0)
nhu.ng f (x0 + 0) 6= f (x0 − 0).
3) Diˆe’m gi´an doa. n kiˆe’u II cu’ a h`am f (x) nˆe´u ta. i diˆe’m x0 mˆo. t trong
f (x) ho˘a. c f (c) khˆong tˆo` n ta. i.
c´ac gi´o.i ha. n lim x→x0+0 H`am f (x) du.o.
.p h`am thu.
lim x→x0−0 .c cho bo.’ i mˆo. t biˆe’u .c go. i l`a h`am so. cˆa´p nˆe´u n´o du.o. th´u.c gia’ i t´ıch lˆa. p nˆen nh`o. mˆo. t sˆo´ h˜u.u ha. n ph´ep t´ınh sˆo´ ho. c v`a c´ac .c hiˆe.n trˆen c´ac h`am so. cˆa´p co. ba’ n. ph´ep ho. Mo. i h`am so. cˆa´p x´ac di.nh trong lˆan cˆa. n cu’ a mˆo. t diˆe’m n`ao d´o l`a
liˆen tu. c ta. i diˆe’m d´o.
Lu.u ´y r˘a`ng h`am khˆong so. cˆa´p c´o thˆe’ c´o gi´an doa. n ta. i nh˜u.ng diˆe’m n´o khˆong x´ac di.nh c˜ung nhu. ta. i nh˜u.ng diˆe’m m`a n´o x´ac di.nh. D˘a. c biˆe.t .c cho bo.’ i nhiˆe`u biˆe’u th´u.c gia’ i t´ıch kh´ac nhau trˆen c´ac l`a nˆe´u h`am du.o. khoa’ ng kh´ac nhau th`ı n´o c´o thˆe’ c´o gi´an doa. n ta. i nh˜u.ng diˆe’m thay dˆo’i biˆe’u th´u.c gia’ i t´ıch.
C ´AC V´I DU. V´ı du. 1. Ch´u.ng minh r˘a`ng h`am f (x) = sin(2x − 3) liˆen tu. c ∀ x ∈ R.
Gia’ i. Ta lˆa´y diˆe’m x0 ∈ R t`uy ´y. X´et hiˆe.u
sin(2x − 3) − sin(2x0 − 3) = 2 cos(x + x0 − 3) sin(x − x0) = α(x).
V`ı | cos(x + x0 − 3)| 6 1 v`a sin(x − x0)| < |x − x0| nˆen khi x → x0 h`am sin(x − x0) l`a h`am vˆo c`ung b´e. T`u. d´o suy r˘a`ng α(x) l`a t´ıch cu’ a h`am bi. ch˘a. n v´o.i vˆo c`ung b´e v`a
sin(2x − 3) = sin(2x0 − 3). N lim x→x0
44 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
√ x + 4 liˆen tu. c ta. i diˆe`m
V´ı du. 2. Ch´u.ng minh r˘a`ng h`am f (x) = x0 = 5.
√ Gia’ i. Ta c´o f (5) = 3. Cho tru.´o.c sˆo´ ε > 0. Theo di.nh ngh˜ıa 1∗ ta .ng mˆodun cu’ a n´o. Ta x + 4 − 3 v`a u.´o.c lu.o. lˆa. p hiˆe.u f (x) − f (5) = c´o
√ √ | < (*) x + 4 − 3| = | |x − 5| 3 |x − 5| x + 4 + 3|
√ Nˆe´u ta cho. n δ = 3ε th`ı v´o.i nh˜u.ng gi´a tri. x m`a |x − 5| < δ = 3ε x + 4 − 3| < ε. T`u. d´o suy r˘a`ng h`am f (x) liˆen tu. c ta. i diˆe’m
√ x liˆen tu. c bˆen pha’ i ta. i
ta s˜e c´o | x0 = 5. N V´ı du. 3. Ch´u.ng minh r˘a`ng h`am f (x) = diˆe’m x0 = 0. √ Gia’ i. Gia’ su.’ cho tru.´o.c sˆo´ ε > 0 t`uy ´y. Bˆa´t d˘a’ ng th´u.c |
√
√ x = 0. N x − 0| < ε tu.o.ng du.o.ng v´o.i bˆa´t d˘a’ ng th´u.c 0 6 x < ε2. Ta lˆa´y δ = ε2. Khi d´o t`u. bˆa´t d˘a’ ng th´u.c 0 6 x < δ suy r˘a`ng x < ε. Diˆe`u d´o c´o ngh˜ıa r˘a`ng lim x→0+0
V´ı du. 4. Ch´u.ng minh r˘a`ng h`am y = x2 liˆen tu. c trˆen to`an tru. c sˆo´.
Gia’ i. Gia’ su.’ x0 ∈ R l`a diˆe’m t`uy ´y trˆen tru. c sˆo´ v`a ε > 0 l`a sˆo´ cho
tru.´o.c t`uy ´y. Ta x´et hiˆe.u
0| = |x + x0||x − x0|
|x2 − x2
.ng n´o. V`ı |x + x0| khˆong bi. ch˘a. n trˆen R nˆen dˆe’ u.´o.c .ng hiˆe.u trˆen ta x´et mˆo. t lˆan cˆa. n n`ao d´o cu’ a x0, ch˘a’ ng ha. n U (x0; 1) =
v`a cˆa` n u.´o.c lu.o. lu.o. (x0 − 1; x0 + 1). V´o.i x ∈ U (x0; 1) ta c´o
|x + x0| = |x − x0 + 2x0| 6 |x − x0| + 2|x0| < 1 + 2|x0|
v`a do d´o
0| < (1 + 2|x0|)|x − x0|.
|x2 − x2
45 7.3. H`am liˆen tu. c
δ = min (cid:16) V`ı δ-lˆan cˆa. n cu’ a diˆe’m x0 cˆa` n pha’ i n˘a`m trong U (x0; 1) nˆen ta lˆa´y ; 1(cid:17) v`a v´o.i |x − x0| < δ = min (cid:16) ; 1(cid:17) ta s˜e ε 1 + 2|x0| ε 1 + 2|x0| c´o
0| < ε. N
|x2 − x2
V´ı du. 5. X´ac di.nh v`a phˆan loa. i diˆe’m gi´an doa. n cu’ a h`am
1 x−1
1 · f (x) = 1 + 2
Gia’ i. H`am d˜a cho x´ac di.nh ∀ x 6= 1. Nhu. vˆa. y diˆe’m gi´an doa. n l`a
diˆe’m x0 = 1.
1
xn−1 (cid:17) l`a d˜ay vˆo
(cid:17) l`a d˜ay vˆo 1 xn − 1
1 xn −1
1 l`a d˜ay vˆo c`ung b´e, t´u.c Nˆe´u (xn) l`a d˜ay hˆo. i tu. dˆe´n 1 v`a xn > 1 th`ı (cid:16) c`ung l´o.n v´o.i mo. i sˆo´ ha. ng dˆe`u du.o.ng. Do d´o (cid:16)1 + 2 c`ung l´o.n. T`u. d´o suy r˘a`ng f (xn) = 1 + 2 f (x) = 0. f (xn) = 0 v`a lim x→1+0
1
xn−1 (cid:17) → 0 (n → ∞) v`a
(cid:17) l`a d˜ay vˆo c`ung l´o.n v´o.i c´ac l`a lim n→∞ Nˆe´u (xn) → 1 v`a xn < 1 th`ı (cid:16) 1 xn − 1
sˆo´ ha. ng dˆe`u ˆam. Do vˆa. y (cid:16)2
1 xn −1
1 → 1 (n → ∞), f (xn) = 1 + 2
f (x) = 1. Do d´o diˆe’m x0 = 1 l`a diˆe’m gi´an doa. n kiˆe’u I. N t´u.c l`a lim x→1−0
V´ı du. 6. X´ac di.nh v`a phˆan loa. i diˆe’m gi´an doa. n cu’ a h`am
khi x < 0 x cos 1 x
f (x) = 0 khi x = 0
cos khi x > 0. 1 x
46 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
Gia’ i. Diˆe’m gi´an doa. n c´o thˆe’ c´o cu’ a h`am l`a x0 = 0. Ta x´et c´ac gi´o.i
ha. n mˆo. t ph´ıa ta. i diˆe’m x0 = 0.
f (x) = 0. Thˆa. t vˆa. y, nˆe´u d˜ay (xn) i) Ta ch´u.ng minh r˘a`ng lim x→0−0
hˆo. i tu. dˆe´n 0 v`a xn < 0 ∀ n th`ı
cos 6 |xn|. 1 xn (cid:12) (cid:12) (cid:12) 0 6 |f (xn)| = |xn|(cid:12) (cid:12) (cid:12)
f (xn) = 0.
n =
n) pha’ i hˆo. i tu. dˆe´n n) = cos 2πn = 1 hˆo. i tu. dˆe´n 1, c`on
1 v`a x0 V`ı |xn| → 0 khi n → ∞ nˆen lim n→∞ ii) H`am d˜a cho khˆong c´o gi´o.i ha. n bˆen pha’ i ta. i diˆe’m x0 = 0. Dˆe’ ch´u.ng minh diˆe`u d´o ta x´et hai d˜ay hˆo. i tu. dˆe´n 0 lˆa. p nˆen t`u. c´ac d˜ay 1 . Nˆe´u nhu. h`am f c´o gi´o.i ha. n sˆo´ du.o.ng xn = 2πn + nπ π 2
bˆen pha’ i ta. i diˆe’m x0 = 0 th`ı hai d˜ay f (xn) v`a f (x0 c`ung mˆo. t gi´o.i ha. n. Thˆe´ nhu.ng f (x0 f (xn) = cos (cid:16) + nπ(cid:17) = 0 hˆo. i tu. dˆe´n 0. π 2 T`u. d´o suy r˘a`ng h`am c´o gi´an doa. n kiˆe’u II ta. i diˆe’m x0 = 0. N
V´ı du. 7. T`ım v`a phˆan loa. i c´ac diˆe’m gi´an doa. n cu’ a c´ac h`am:
1) y = (signx)2; 2) y = [x]
Gia’ i 1) T`u. di.nh ngh˜ıa h`am signx suy r˘a`ng
1, x 6= 0 (signx)2 = 0, x = 0.
.c. .ng dˆo` T`u. d´o suy r˘a`ng h`am y = (signx)2 liˆen tu. c ∀ x 6= 0 (h˜ay du. thi. cu’ a h`am) v`a ta. i diˆe’m x0 = 0 ta c´o y(0 − 0) = y(0 + 0) 6= y(0). Diˆe`u d´o c´o ngh˜ıa r˘a`ng x0 = 0 l`a diˆe’m gi´an doa. n khu.’ du.o.
2) Gia’ su.’ n ∈ Z. Nˆe´u n − 1 6 x < n th`ı [x] = n − 1, nˆe´u .ng dˆo` thi. cu’ a h`am phˆa` n nguyˆen n 6 x < n + 1 th`ı [x] = n (h˜ay du. [x]). Nˆe´u x0 6∈ Z th`ı tˆo` n ta. i lˆan cˆa. n cu’ a diˆe’m x0 (khˆong ch´u.a c´ac sˆo´
47 7.3. H`am liˆen tu. c
. liˆen tu. c v`a phˆan loa. i diˆe’m gi´an doa. n cu’ a c´ac nguyˆen) sao cho ta. i d´o h`am b˘a`ng h˘a`ng sˆo´. Do vˆa. y n´o liˆen tu. c ta. i x0. Nˆe´u x0 = n l`a sˆo´ nguyˆen th`ı [n − 0] = n − 1, [n + 0] = n. T`u. d´o suy r˘a`ng x0 = n l`a diˆe’m gi´an doa. n kiˆe’u I. N V´ı du. 8. Kha’ o s´at su. h`am
nˆe´u x 6 1 x 1) f (x) = 2) , f (x) = e− 1 x , 3) f (x) = x2 x lnx nˆe´u x > 1.
Gia’ i 1) H`am f (x) = x nˆe´u x 6= 0 v`a khˆong x´ac di.nh khi x = 0. V`ı ∀ a x = a nˆen khi a 6= 0: ta c´o lim x→a
f (x) = a = f (a) lim x→a
v`a do vˆa. y h`am f (x) liˆen tu. c ∀ x 6= 0. Ta. i diˆe’m x = 0 ta c´o gi´an doa. n khu.’ du.o. .c v`ı tˆo` n ta. i
x = 0. lim x→0 f (x) = lim x→0
x l`a h`am so. cˆa´p v`ı n´o l`a ho.
2) H`am f (x) = e− 1
.p cu’ a c´ac h`am y = −x−1 v`a f = ey. Hiˆe’n nhiˆen l`a h`am f (x) x´ac di.nh ∀ x 6= 0 v`a do d´o n´o liˆen tu. c ∀ x 6= 0. V`ı h`am f (x) x´ac di.nh trong lˆan cˆa. n diˆe’m x = 0 v`a khˆong x´ac di.nh ta. i ch´ınh diˆe’m x = 0 nˆen diˆe’m x = 0 l`a diˆe’m gi´an doa. n. Ta t´ınh f (0 + 0) v`a f (0 − 0).
Ta x´et d˜ay vˆo c`ung b´e t`uy ´y (xn) sao cho xn > 0 ∀ n. V`ı (cid:16)− x = 0. e− 1 e− 1 (cid:17) = −∞ nˆen lim x→∞ 1 xn
xn = 0. T`u. d´o suy r˘a`ng lim x→0+0 0 < 0 ∀ n. V`ı x = +∞
n) sao cho x0 − 1 n = +∞. Do d´o lim x0 e x→0−0
(cid:16) − e− 1
lim x→∞ Bˆay gi`o. ta x´et d˜ay vˆo c`ung b´e bˆa´t k`y (x0 1 (cid:17) = +∞ nˆen lim lim x0 x→0 n→∞ n t´u.c l`a f (0 − 0) = +∞.
Nhu. vˆa. y gi´o.i ha. n bˆen tr´ai cu’ a h`am f (x) ta. i diˆe’m x = 0 khˆong tˆo` n
ta. i do d´o diˆe’m x = 0 l`a diˆe’m gi´an doa. n kiˆe’u II.
48 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
3) Ta ch´u.ng minh r˘a`ng f (x) liˆen tu. c ta. i diˆe’m x = a 6= 1. Ta lˆa´y ε < |a − 1|, ε > 0. Khi d´o ε-lˆan cˆa. n cu’ a diˆe’m x = a khˆong ch´u.a diˆe’m x = 1 nˆe´u ε < |a − 1|. Trong ε-lˆan cˆa. n n`ay h`am f (x) ho˘a. c tr`ung v´o.i h`am ϕ(x) = x nˆe´u a < 1 ho˘a. c tr`ung v´o.i h`am ϕ(x) = lnx nˆe´u a > 1. V`ı c´ac h`am so. cˆa´p co. ba’ n n`ay liˆen tu. c ta. i diˆe’m x = a nˆen h`am f (x) liˆen tu. c ta. i diˆe’m x = a 6= 1.
Ta kha’ o s´at t´ınh liˆen tu. c cu’ a h`am f (x) ta. i diˆe’m x = a = 1. Dˆe’ l`am viˆe.c d´o ta cˆa` n t´ınh c´ac gi´o.i ha. n mˆo. t ph´ıa cu’ a f (x) ta. i diˆe’m x = a = 1. Ta c´o
lnx = 0, f (x) = lim x→1+0
x = 1. f (1 + 0) = lim x→1+0 f (1 − 0) = lim x→1−0 f (x) = lim x→1−0 x = lim x→1
Nhu. vˆa. y f (1 + 0) 6= f (1 − 0) v`a do d´o h`am f (x) c´o gi´an doa. n kiˆe’u
I ta. i x = a = 1.
B `AI T ˆA. P
1. f (x) = ; ta. i (DS. H`am x´ac di.nh v`a liˆen tu. c ∀ x 6= Kha’ o s´at t´ınh liˆen tu. c v`a phˆan loa. i diˆe’m gi´an doa. n cu’ a h`am 3 2 |2x − 3| 2x − 3
x0 = h`am c´o gi´an doa. n kiˆe’u I) 3 2
nˆe´u x 6= 0 2. f (x) = 1 x 1 nˆe´u x = 0. (DS. H`am liˆen tu. c ∀ x ∈ R)
3. C´o tˆo` n ta. i hay khˆong gi´a tri. a dˆe’ h`am f (x) liˆen tu. c ta. i x0 nˆe´u:
4 · 3x nˆe´u x < 0 1) f (x) = 2a + x khi x > 0. (DS. H`am f liˆen tu. c ∀ x ∈ R nˆe´u a = 2)
49 7.3. H`am liˆen tu. c
x sin , x 6= 0; 1 x 2) f (x) = . a, x = 0, x0 = 0.
(DS. a = 0)
, x 6= −1 3) f (x) = 1 + x 1 + x3 a, x = −1, x0 = −1.
(DS. a =
cos x, x 6 0; 1 ) 3 4) f (x) = a(x − 1), x > 0; x0 = 0.
4. f (x) = (DS. a = −1) | sin x| sin x (DS. H`am c´o gi´an doa. n ta. i x = kπ, k ∈ Z v`ı:
nˆe´u sin x > 0 1 f (x) = −1 nˆe´u sin x < 0)
5. f (x) = E(x) − E(−x)
(DS. H`am c´o gi´an doa. n khu.’ du.o. .c ta. i x = n, x ∈ Z v`ı:
−1 nˆe´u x = n f (x) = nˆe´u x 6= n.) 0
e1/x khi x 6= 0 6. f (x) = khi x = 0. 0 (DS. Ta. i diˆe’m x = 0 h`am c´o gi´an doa. n kiˆe’u II; f (−0) = 0, f (+0) = ∞)
T`ım diˆe’m gi´an doa. n v`a t´ınh bu.´o.c nha’ y cu’ a c´ac h`am:
7. f (x) = x + x + 2 |x + 2|
(DS. x = −2 l`a diˆe’m gi´an doa. n kiˆe’u I, δ(−2) = 2)
50 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
8. f (x) = 2|x − 1| x2 − x3
(DS. x = 0 l`a diˆe’m gi´an doa. n kiˆe’u II, x = 1 l`a diˆe’m gi´an doa. n kiˆe’u I, δ(1) = −4)
H˜ay bˆo’ sung c´ac h`am sau dˆay ta. i diˆe’m x = 0 dˆe’ ch´ung tro.’ th`anh
liˆen tu. c
(DS. f (0) = 1) 9. f (x) =
tgx x √
) 10. f (x) = (DS. f (0) = 1 2
11. f (x) = (DS. f (0) = 2) 1 + x − 1 x sin2 x 1 − cos x
12. Hiˆe.u cu’ a c´ac gi´o.i ha. n mˆo. t ph´ıa cu’ a h`am f (x):
x→x0+0
x→x0−0
d = lim f (x) − lim f (x)
.c go. i l`a bu.´o.c nha’ y cu’ a h`am f (x) ta. i diˆe’m x0. T`ım diˆe’m gi´an doa. n du.o. v`a bu.´o.c nha’ y cu’ a h`am f (x) nˆe´u:
− x2 nˆe´u x 6 2, 1 2 1) f (x) = x nˆe´u x > 2. (DS. x0 = 2 l`a diˆe’m gi´an doa. n kiˆe’u I; d = 4)
x nˆe´u 0 6 x 6 1; √ 2
2) f (x) = 4 − 2x nˆe´u 1 < x 6 2, 5;
2x − 7 nˆe´u 2, 5 6 x < +∞. (DS. x0 = 2, 5 l`a diˆe’m gi´an doa. n kiˆe’u I; d = −1)
3) f (x) =
2x + 5 nˆe´u − ∞ < x < −1, 1 x nˆe´u − 1 6 x < +∞. (DS. x0 = 0 l`a diˆe’m gi´an doa. n kiˆe’u II; diˆe’m x0 = −1 l`a diˆe’m gi´an
doa. n kiˆe’u I, d = −4)
51
7.4. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n 7.4 Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u
biˆe´n
1. Gia’ su.’ u = f (M) = f (x, y) x´ac di.nh trˆen tˆa. p ho. .p D. Gia’ su.’ M0(x0, y0) l`a diˆe’m cˆo´ di.nh n`ao d´o cu’ a m˘a. t ph˘a’ ng v`a x → x0, y → y0, khi d´o diˆe’m M(x, y) → M0(x0, y0). Diˆe`u n`ay tu.o.ng du.o.ng v´o.i khoa’ ng c´ach ρ(M, M0) gi˜u.a hai diˆe’m M v`a M0 dˆa` n dˆe´n 0. Ta lu.u ´y r˘a`ng
ρ(M, M0) = [(x − x0)2 + (y − y0)2]1/2.
.c go. i l`a gi´o.i ha. n cu’ a h`am f (M) khi M → M0 (hay ta. i
Ta c´o c´ac di.nh ngh˜ıa sau dˆay: i) Di.nh ngh˜ıa gi´o.i ha. n (theo Cauchy) Sˆo´ b du.o. diˆe’m M0) nˆe´u
∀ ε > 0, ∃ δ = δ(ε) > 0 : ∀ M ∈ {D : 0 < ρ(M, M0) < δ(ε)} ⇒ |f (M) − b| < ε.
ii) Di.nh ngh˜ıa gi´o.i ha. n (theo Heine) Sˆo´ b du.o.
.c go. i l`a gi´o.i ha. n cu’ a h`am f (M) ta. i diˆe’m M0 nˆe´u dˆo´i v´o.i d˜ay diˆe’m {Mn} bˆa´t k`y hˆo. i tu. dˆe´n M0 sao cho Mn ∈ D, Mn 6= M0 ∀ n ∈ N th`ı d˜ay c´ac gi´a tri. tu.o.ng ´u.ng cu’ a h`am {f (Mn)} hˆo. i tu. dˆe´n b.
K´y hiˆe.u:
i) f (M) = b, ho˘a. c
f (x, y) = b ii)
lim M →M0 lim x → x0 y → y0
Hai di.nh ngh˜ıa gi´o.i ha. n trˆen dˆay tu.o.ng du.o.ng v´o.i nhau.
Ch´u ´y. Ta nhˆa´n ma. nh r˘a`ng theo di.nh ngh˜ıa, gi´o.i ha. n cu’ a h`am khˆong phu. thuˆo. c v`ao phu.o.ng M dˆa` n t´o.i M0. Do d´o nˆe´u M → M0 theo c´ac hu.´o.ng kh´ac nhau m`a f (M) dˆa` n dˆe´n c´ac gi´a tri. kh´ac nhau th`ı khi M → M0 h`am f (M) khˆong c´o gi´o.i ha. n.
52 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´
iii) Sˆo´ b du.o. .c go. i l`a gi´o.i ha. n cu’ a h`am f (M) khi M → ∞ nˆe´u
∀ ε > 0, ∃ R > 0 : ∀ M ∈ {D : ρ(M, 0) > R} ⇒ |f (M) − b| < ε.
Dˆo´i v´o.i h`am nhiˆe`u biˆe´n, c`ung v´o.i gi´o.i ha. n thˆong thu.`o.ng d˜a nˆeu o.’ trˆen (gi´o.i ha. n k´ep !), ngu.`o.i ta c`on x´et gi´o.i ha. n l˘a. p. Ta s˜e x´et kh´ai niˆe.m n`ay cho h`am hai biˆe´n u = f (M) = f (x, y).
Gia’ su.’ u = f (x, y) x´ac di.nh trong h`ınh ch˜u. nhˆa. t
Q = {(x, y) : |x − x0| < d1, |y − y0| < d2}
c´o thˆe’ tr`u. ra ch´ınh c´ac diˆe’m x = x0, y = y0. Khi cˆo´ di.nh mˆo. t gi´a tri. y th`ı h`am f (x, y) tro.’ th`anh h`am mˆo. t biˆe´n. Gia’ su.’ dˆo´i v´o.i gi´a tri. cˆo´ di.nh y bˆa´t k`y tho’a m˜an diˆe`u kiˆe.n 0 < |y − y0| < d2 tˆo` n ta. i gi´o.i ha. n
f (x, y) = ϕ(y).
lim x→x0 y cˆo´ di.nh
Tiˆe´p theo, gia’ su.’ ϕ(y) = b tˆo` n ta. i. Khi d´o ngu.`o.i ta n´oi r˘a`ng lim y→y0
tˆo` n ta. i gi´o.i ha. n l˘a. p cu’ a h`am f (x, y) ta. i diˆe’m M0(x0, y0) v`a viˆe´t
f (x, y) = b, lim y→y0
., ta lim x→x0 f (x, y) go. i l`a gi´o.i ha. n trong. Tu.o.ng tu. trong d´o gi´o.i ha. n
lim
x→x0
y cˆo´ di.nh
0<|y−y0| f (x, y) lim
y→y0
x cˆo´ di.nh
0<|x−x0| l`a gi´o.i ha. n trong. .c thˆe’ hiˆe.n Mˆo´i quan hˆe. gi˜u.a gi´o.i ha. n k´ep v`a c´ac gi´o.i ha. n l˘a. p du.o. trong di.nh l´y sau dˆay: 53 7.4. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n Gia’ su.’ ta. i diˆe’m M0(x0, y0) gi´o.i ha. n k´ep v`a c´ac gi´o.i ha. n trong cu’ a
c´ac gi´o.i ha. n l˘a. p cu’ a h`am tˆo` n ta. i. Khi d´o c´ac gi´o.i ha. n l˘a. p tˆo` n ta. i v`a f (x, y). lim
x→x0 lim
y→y0 f (x, y) = lim
y→y0 lim
x→x0 = lim
x→x0
y→y0 . trong c´ac gi´o.i T`u. di.nh l´y n`ay ta thˆa´y r˘a`ng viˆe.c thay dˆo’i th´u. tu. ha. n khˆong pha’ i bao gi`o. c˜ung du.o. .c ph´ep.
Dˆo´i v´o.i h`am nhiˆe`u biˆe´n ta c˜ung c´o nh˜u.ng di.nh l´y vˆe` c´ac t´ınh chˆa´t
. c´ac di.nh l´y vˆe` gi´o.i ha. n cu’ a h`am mˆo. t sˆo´ ho. c cu’ a gi´o.i ha. n tu.o.ng tu.
biˆe´n.
2. T`u. kh´ai niˆe.m gi´o.i ha. n ta s˜e tr`ınh b`ay kh´ai niˆe.m vˆe` t´ınh liˆen tu. c
cu’ a h`am nhiˆe`u biˆe´n. .c go. i l`a liˆen tu. c ta. i diˆe’m M0 nˆe´u: H`am u = f (M) du.o.
i) f (M) x´ac di.nh ta. i ch´ınh diˆe’m M0 c˜ung nhu. trong mˆo. t lˆan cˆa. n f (M) tˆo` n ta. i. n`ao d´o cu’ a diˆe’m M0.
ii) Gi´o.i ha. n lim
M →M0
iii) f (M) = f (M0). .c di.nh ngh˜ıa go. i l`a su. lim
M →M0
. liˆen tu. c v`u.a du.o. .p
. liˆen tu. c theo tˆa. p ho. Su.
biˆe´n sˆo´. H`am f (M) liˆen tu. c trong miˆe`n D nˆe´u n´o liˆen tu. c ta. i mo. i diˆe’m cu’ a miˆe`n d´o. Diˆe’m M0 du.o. .c go. i l`a diˆe’m gi´an doa. n cu’ a h`am f (M) nˆe´u dˆo´i v´o.i
diˆe’m M0 c´o ´ıt nhˆa´t mˆo. t trong ba diˆe`u kiˆe.n trong di.nh ngh˜ıa liˆen tu. c
khˆong tho’a m˜an. Diˆe’m gi´an doa. n cu’ a h`am nhiˆe`u biˆe´n c´o thˆe’ l`a nh˜u.ng
diˆe’m cˆo lˆa. p, v`a c˜ung c´o thˆe’ l`a ca’ mˆo. t du.`o.ng (du.`o.ng gi´an doa. n).
Nˆe´u h`am f (x, y) liˆen tu. c ta. i diˆe’m M0(x0, y0) theo tˆa. p ho. .p biˆe´n sˆo´
.c la. i l`a khˆong th`ı n´o liˆen tu. c theo t`u.ng biˆe´n sˆo´. Diˆe`u kh˘a’ ng di.nh ngu.o.
d´ung. C˜ung nhu. dˆo´i v´o.i h`am mˆo. t biˆe´n, tˆo’ng, hiˆe.u v`a t´ıch c´ac h`am liˆen
tu. c hai biˆe´n ta. i diˆe’m M0 l`a h`am liˆen tu. c ta. i diˆe’m d´o; thu.o.ng cu’ a hai
h`am liˆen tu. c ta. i M0 c˜ung l`a h`am liˆen tu. c ta. i M0 nˆe´u ta. i diˆe’m M0 h`am 54 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´ .p vˆa˜n .p n`ay.
. nhu. trˆen ta c´o thˆe’ tr`ınh b`ay c´ac kh´ai niˆe.m co. mˆa˜u sˆo´ kh´ac 0. Ngo`ai ra, di.nh l´y vˆe` t´ınh liˆen tu. c cu’ a h`am ho.
d´ung trong tru.`o.ng ho.
Nhˆa. n x´et. Tu.o.ng tu.
ba’ n liˆen quan dˆe´n gi´o.i ha. n v`a liˆen tu. c cu’ a h`am ba biˆe´n,... C ´AC V´I DU. V´ı du. 1. Ch´u.ng minh r˘a`ng h`am sin l`a vˆo c`ung b´e ta. i diˆe’m O(0, 0). 1
y 1
f (x, y) = (x + y) sin
x
Gia’ i. Theo di.nh ngh˜ıa vˆo c`ung b´e (tu.o.ng tu. . nhu. dˆo´i v´o.i h`am mˆo. t biˆe´n) ta cˆa` n ch´u.ng minh r˘a`ng f (x, y) = 0. lim
x→0
y→0 Ta ´ap du. ng di.nh ngh˜ıa gi´o.i ha. n theo Cauchy. Ta cho sˆo´ ε > 0 t`uy . Khi d´o nˆe´u ´y v`a d˘a. t δ = ε
2 ρ(cid:2)M(x, y), O(0, 0)(cid:3) = px2 + y2 < δ th`ı |x| < δ, |y| < δ. Do d´o sin 6 |x| + |y| < 2δ = ε. (x + y) sin 1
x 1
y (cid:12)
(cid:12)
(cid:12) |f (x, y) − 0| = (cid:12)
(cid:12)
(cid:12) Diˆe`u d´o ch´u.ng to’ r˘a`ng f (x, y) = 0. lim
x→0
y→0 V´ı du. 2. T´ınh c´ac gi´o.i ha. n sau dˆay: , 2
x2 + xy , (cid:0)1 + xy(cid:1) px2 + (y − x)2 + 1 − 1
x2 + (y − 2)2 2) lim
x→0
y→2 1) lim
x→0
y→2 x4 + y4
x2 + y2 . 3) lim
x→0
y→0 55 7.4. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n Gia’ i. 1) Ta biˆe’u diˆe˜n h`am du.´o.i dˆa´u gi´o.i ha. n du.´o.i da. ng 1
xy i 2y
x + y . h(cid:0)1 + xy(cid:1) ! V`ı t = xy → 0 khi nˆen x → 0
y → 0 1
t = e. (cid:0)1 + t(cid:1) (cid:0)1 + xy(cid:1) 1
xy = lim
t→0 lim
x→0
y→2 = 2 (theo di.nh l´y thˆong thu.`o.ng vˆe` gi´o.i ha. n Tiˆe´p theo v`ı lim
x→0
y→2 2
x + y
cu’ a thu.o.ng), do d´o gi´o.i ha. n cˆa` n t`ım b˘a`ng e2. 2) Ta t`ım gi´o.i ha. n v´o.i diˆe`u kiˆe.n M(x, y) → M0(0, 2). Khoa’ ng c´ach gi˜u.a hai diˆe’m M v`a M0 b˘a`ng ρ = px2 + (y − 2)2 . Do d´o f (x, y) = lim
ρ→0 = lim
ρ→0 pρ2 + 1 − 1
ρ2 (ρ2 + 1) − 1
ρ2(pρ2 + 1 + 1) lim
x→0
y→2 · = = lim
ρ→0 1
2 1
pρ2 + 1 + 1 .c ta c´o x = ρ cos ϕ, y = ρ sin ϕ. Ta c´o 3) Chuyˆe’n sang to. a dˆo. cu. = ρ2(cos4 ϕ + sin4 ϕ). x4 + y4
x2 + y2 = ρ4(cos4 ϕ + sin4 ϕ)
ρ2(cos2 ϕ + sin2 ϕ) V`ı cos4 ϕ + sin4 ϕ 6 2 nˆen ρ→0 ρ2(cos4 ϕ + sin4 ϕ) = 0. x4 + y4
x2 + y2 = lim lim
x→0
y→0 56 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´ V´ı du. 3. 1) Ch´u.ng minh r˘a`ng h`am f1(x, y) = x − y
x + y khˆong c´o gi´o.i ha. n ta. i diˆe’m (0, 0). 2) H`am f2(x, y) = xy
x2 + y2 c´o gi´o.i ha. n ta. i diˆe’m (0, 0) hay khˆong ? Gia’ i. 1) H`am f1(x, y) x´ac di.nh kh˘a´p no.i ngoa. i tr`u. du.`o.ng th˘a’ ng
x + y = 0. Ta ch´u.ng minh r˘a`ng h`am khˆong c´o gi´o.i ha. n ta. i (0, 0). Ta
lˆa´y hai d˜ay diˆe’m hˆo. i tu. dˆe´n diˆe’m (0, 0): , 0(cid:17) → (0, 0), n → ∞, Mn = (cid:16) M 0 (cid:17) → (0, 0), n → ∞. 1
n
n = (cid:16)0, 1
n .c
Khi d´o thu du.o. − 0 = 1; lim
n→∞ f1(Mn) = lim
n→∞ + 0 1
n
1
n n) = lim
n→∞ 0 − = −1. f1(M 0 lim
n→∞ 0 + 1
n
1
n Nhu. vˆa. y hai d˜ay diˆe’m kh´ac nhau c`ung hˆo. i tu. dˆe´n diˆe’m (0, 0) nhu.ng
hai d˜ay gi´a tri. tu.o.ng ´u.ng cu’ a h`am khˆong c´o c`ung gi´o.i ha. n. Do d´o
theo di.nh ngh˜ıa h`am khˆong c´o gi´o.i ha. n ta. i (0, 0). 2) Gia’ su.’ diˆe’m M(x, y) dˆa` n dˆe´n diˆe’m (0, 0) theo du.`o.ng th˘a’ ng y = kx qua gˆo´c to. a dˆo. . Khi d´o ta c´o · = = lim
x→0 xy
x2 + y2 kx2
x2 + k2x2 k
1 + k2 lim
x→0
y→0
(y=kx) 57 7.4. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n Nhu. vˆa. y khi dˆa` n dˆe´n diˆe’m (0, 0) theo c´ac du.`o.ng th˘a’ ng kh´ac nhau
(tu.o.ng ´u.ng v´o.i c´ac gi´a tri. k kh´ac nhau) ta thu du.o.
.c c´ac gi´a tri. gi´o.i
ha. n kh´ac nhau, t´u.c l`a h`am d˜a cho khˆong c´o gi´o.i ha. n ta. i (0, 0). N 1) f (x, y) = 2) f (x, y) = 3) f (x, y) = V´ı du. 4. Kha’ o s´at t´ınh liˆen tu. c cu’ a c´ac h`am
x2 + 2xy + 5
y2 − 2x + 1
1
x2 + y2 − z
x + y
x3 + y3 Gia’ i. 1) Diˆe`u kiˆe.n liˆen tu. c cu’ a h`am d˜a cho bi. vi pha. m ta. i nh˜u.ng
diˆe’m cu’ a m˘a. t ph˘a’ ng R2 m`a to. a dˆo. cu’ a ch´ung tho’a m˜an phu.o.ng tr`ınh
y2 − 2x + 1 = 0. D´o l`a phu.o.ng tr`ınh du.`o.ng parabˆon v´o.i dı’nh ta. i diˆe’m
1
, 0(cid:17). Nhu. vˆa. y c´ac diˆe’m cu’ a parabˆon n`ay l`a nh˜u.ng diˆe’m gi´an doa. n
(cid:16)
2
- d´o l`a du.`o.ng gi´an doa. n cu’ a h`am. Nh˜u.ng diˆe’m cu’ a m˘a. t ph˘a’ ng R2
khˆong thuˆo. c parabˆon d´o l`a nh˜u.ng diˆe’m liˆen tu. c. 2) H`am d˜a cho liˆen tu. c ta. i mo. i diˆe’m cu’ a khˆong gian R3 m`a to. a dˆo.
cu’ a ch´ung tho’a m˜an diˆe`u kiˆe.n x2 + y2 − z 6= 0. D´o l`a phu.o.ng tr`ınh
m˘a. t paraboloit tr`on xoay. Trong tru.`o.ng ho.
.p n`ay m˘a. t paraboloit l`a
m˘a. t gi´an doa. n cu’ a h`am. 3) V`ı tu.’ sˆo´ v`a mˆa˜u sˆo´ l`a nh˜u.ng h`am liˆen tu. c nˆen thu.o.ng l`a h`am
liˆen tu. c ta. i nh˜u.ng diˆe’m m`a mˆa˜u sˆo´ x3 + y3 6= 0. H`am c´o gi´an doa. n ta. i
nh˜u.ng diˆe’m m`a x3 + y3 = 0 hay y = −x. Ngh˜ıa l`a h`am c´o gi´an doa. n
trˆen du.`o.ng th˘a’ ng y = −x. Gia’ su.’ x0 6= 0, y0 6= 0. Khi d´o · = x + y
x3 + y3 1
x2 − xy + y2 1
0 − x0y0 + y2
x2
0 lim
x→x0
y→y0 = lim
x→x0
y→y0 T`u. d´o suy ra r˘a`ng c´ac diˆe’m cu’ a du.`o.ng th˘a’ ng y = x (x 6= 0) l`a 58 Chu.o.ng 7. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am sˆo´ .c. V`ı nhu. .ng diˆe’m gi´an doa. n khu.’ du.o. = +∞ x + y
x3 + y3 1
x2 − xy + y2 lim
x→0
y→0 = lim
x→0
y→0 nˆen diˆe’m O(0, 0) l`a diˆe’m gi´an doa. n vˆo c`ung. B `AI T ˆA. P Trong c´ac b`ai to´an sau dˆay (1-10) h˜ay t`ım miˆe`n x´ac di.nh cu’ a c´ac h`am nˆe´u: (DS. |y| 6 |x|) 1. w = px2 − y2.
√ xy. 2. w = (DS. x > 0, y > 0 ho˘a. c x 6 0, y 6 0) (DS. x2 + y2 6 a2) 3. w = pa2 − x2 − y2. 4. w = . (DS. x2 + y2 > a2) 1
px2 + y2 − a2 r − 1 − . (DS. + 6 1) 5. w = x2
a2 y2
b2 y2
b2 x2
a2
(DS. x2 + y2 − z2 < −1) √ + xy. 7. w = arcsin (DS. Hai nu.’ a b˘ang vˆo ha. n th˘a’ ng d´u.ng 6. w = ln(z2 − x2 − y2 − 1).
x
2 {0 6 x 6 2, 0 6 y < +∞} v`a {−2 6 x 6 0, −∞ < y 6 0}) 8. w = px2 + y2 − 1 + ln(4 − x2 − y2).
(DS. V`anh tr`on 1 6 x2 + y2 < 4) .p c´ac v`anh dˆo` ng tˆam 9. w = psin π(x2 + y2). (DS. Tˆa. p ho.
0 6 x2 + y2 6 1; 2 6 x2 + y2 6 3; . . . ) 10. w = pln(1 + z − x2 − y2). (DS. Phˆa` n trong cu’ a mˆa. t paraboloid z = x2 + y2 − 1).
Trong c´ac b`ai to´an sau dˆay (11-18) h˜ay t´ınh c´ac gi´o.i ha. n cu’ a h`am 59 7.4. Gi´o.i ha. n v`a liˆen tu. c cu’ a h`am nhiˆe`u biˆe´n . 11. (DS. 1) sin xy
xy lim
x→0
y→0 . (DS. 0) 12. sin xy
x lim
x→0
y→0 √ . (DS. 2) 13. xy
xy + 1 − 1 lim
x→0
y→0 14. . (DS. 2) x2 + y2
px2 + y2 + 1 − 1 lim
x→0
y→0 Chı’ dˆa˜n. Su.’ du. ng khoa’ ng c´ach ρ = px2 + y2 ho˘a. c nhˆan - chia .p v´o.i mˆa˜u sˆo´. .ng liˆen ho. v´o.i da. i lu.o. 15. (DS. e3) y
x2y + xy2 . (cid:0)1 + xy2(cid:1) lim
x→0
y→3 16. . (DS. 0) x2y
x2 + y2 lim
x→0
y→0 ) . (DS. 17. p(x2 + (y − 5)2 + 1 − 1
x2 + (y − 5)2 1
2 lim
x→0
y→5 18. . (DS. 2). tg(2xy)
x2y lim
x→1
y→0 8.1 D- a. o h`am . . . . . . . . . . . . . . . . . . . . . 61 8.1.1 D- a.o h`am cˆa´p 1 . . . . . . . . . . . . . . . . 61 8.1.2 D- a.o h`am cˆa´p cao . . . . . . . . . . . . . . . 62 8.2 Vi phˆan . . . . . . . . . . . . . . . . . . . . . 75 8.2.1 Vi phˆan cˆa´p 1 . . . . . . . . . . . . . . . . . 75 8.2.2 Vi phˆan cˆa´p cao . . . . . . . . . . . . . . . 77 8.3 C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi. Quy t˘a´c l’Hospital. Cˆong th´u.c Taylor . . . . . . 84 . . . . . 84 8.3.1 C´ac d i.nh l´y co. ba’n vˆe` h`am kha’ vi
8.3.2 Khu.’ c´ac da. ng vˆo di.nh. Quy t˘a´c Lˆopitan (L’Hospitale) . . . . . . . . . . . . . . . . . 88 8.3.3 Cˆong th´u.c Taylor . . . . . . . . . . . . . . . 96 61 8.1. D- a. o h`am Theo di.nh ngh˜ıa: Nˆe´u tˆo` n ta. i gi´o.i ha. n h˜u.u ha. n lim
∆x→0 f (x0 + ∆x) − f (x0)
∆x .c go. i l`a da. o h`am cu’ a h`am f (x) ta. i khi ∆x → 0 th`ı gi´o.i ha. n d´o du.o.
diˆe’m x0 v`a du.o. .c chı’ bo.’ i mˆo. t trong c´ac k´y hiˆe.u: ≡ ≡ f (x) ≡ f 0(x) ≡ y0. lim
∆x→0 f (x0 + ∆x) − f (x0)
∆x dy
dx d
dx .ng Da. i lu.o. ∆x→0+0 +(x0) = f 0(x0 + 0) = lim
f 0
∆x→0
∆x>0 = lim ∆y
∆x ∆y
∆x v`a ∆x→0−0 −(x0) = f 0(x0 − 0) = lim
f 0
∆x→0
∆x<0 = lim ∆y
∆x ∆y
∆x .c go. i l`a da. o h`am bˆen pha’ i v`a da. o h`am bˆen tr´ai cu’ a h`am y = f (x) du.o.
ta. i diˆe’m x0 nˆe´u c´ac gi´o.i ha. n d˜a nˆeu tˆo` n ta. i. Su.’ du. ng kh´ai niˆe.m gi´o.i ha. n mˆo. t ph´ıa ta c´o: D- i.nh l´y 8.1.1. H`am y = f (x) c´o da. o h`am ta. i diˆe’m x khi v`a chı’ khi
c´ac da. o h`am mˆo. t ph´ıa tˆo` n ta. i v`a b˘a`ng nhau: f 0(x + 0) = f 0(x − 0) = f 0(x). H`am f (x) kha’ vi nˆe´u n´o c´o da. o h`am f 0(x) h˜u.u ha. n. H`am f (x) kha’
vi liˆen tu. c nˆe´u da. o h`am f 0(x) tˆo` n ta. i v`a liˆen tu. c. Nˆe´u h`am f (x) kha’
vi th`ı n´o liˆen tu. c. Diˆe`u kh˘a’ ng di.nh ngu.o. .c la. i l`a khˆong d´ung. 62 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 8.1.2 D- a. o h`am cˆa´p cao
Da. o h`am f 0(x) du.o.
Da. o h`am cu’ a f 0(x) du.o.
hai) cu’ a h`am f (x) v`a du.o.
f 00(x) du.o.
v`a du.o. .c go. i l`a da. o h`am cˆa´p 1 (hay da. o h`am bˆa. c nhˆa´t).
.c go. i l`a da. o h`am cˆa´p hai (hay da. o h`am th´u.
.c k´y hiˆe.u l`a y00 hay f 00(x). Da. o h`am cu’ a
.c go. i l`a da. o h`am cˆa´p 3 (hay da. o h`am th´u. ba) cu’ a h`am f (x) .c k´y hiˆe.u y000 hay f 000(x) (hay y(3), f (3)(x) v.v... Ta c´o ba’ ng da. o h`am cu’ a c´ac h`am so. cˆa´p co. ba’ n f (x) f 0(x) f (n)(x) xa axa−1 ex
ax lnx (−1)n−1(n − 1)! , x > 0 , x > 0 (−1)n−1(n − 1)! logax ex
axlna
1
x
1
xlna (cid:17) sin x cos x sin (cid:16)x + a(a − 1)(a − 2) · · · (a − n + 1)xa−n,
x > 0
ex
ax(lna)n
1
xn
1
xnlna
nπ
2 63 8.1. D- a. o h`am f (n)(x) f (x) f 0(x) (cid:17) cos (cid:16)x + cos x − sin x nπ
2 tgx − cotgx 1
cos2 x
1
sin2 x √ , |x| < 1 arc sin x √ − , |x| < 1 arccosx arctgx − arccotgx 1
1 − x2
1
1 − x2
1
1 + x2
1
1 + x2 .a trˆen c´ac quy t˘a´c sau dˆay. Viˆe.c t´ınh da. o h`am du.o. 1+ v. [u + v] = u + .c du.
d
dx 2+ (αu) = α , α ∈ R. 3+ (uv) = v + u . dv
dx (cid:16) (cid:17) = (cid:17), v 6= 0. − u (cid:16)v 4+ d
dx
du
dx
du
dx
1
v2 u
v du
dx dv
dx · 5+ f [u(x)] = .p). (da. o h`am cu’ a h`am ho. d
dx
d
dx
d
dx
d
dx
d
dx du
dx x 6= 0 th`ı ≡ y0 .c x = x(y) v`a df
du
6+ Nˆe´u h`am y = y(x) c´o h`am ngu.o. dy
dx y = · ≡ x0 dx
dy 1
y0
x 64 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n y 6= 0 th`ı 7+ Nˆe´u h`am y = y(x) du.o. .c cho du.´o.i da. ng ˆa’n bo.’ i hˆe. th´u.c kha’ vi F (x, y) = 0 v`a F 0 y l`a da. o h`am theo biˆe´n tu.o.ng ´u.ng cu’ a h`am F (x, y) x v`a F 0
trong d´o F 0
khi xem biˆe´n kia khˆong dˆo’i. = − dy
dx F 0
x
F 0
y 8+ Nˆe´u h`am y = y(x) du.o. .c cho du.´o.i da. ng tham sˆo´ x = x(t), y = y(t) (x0(t) 6= 0) th`ı · = dy
dx y0(t)
x0(t) 9+ (αu + βv) = α + β ; dn
dxn dnu
dxn dnv
dxn n
X
k=0 uv = u v (quy t˘a´c Leibniz). C k
n dn
dxn dn−k
dxn−k dk
dxk Nhˆa. n x´et. 1) Khi t´ınh da. o h`am cu’ a mˆo. t biˆe’u th´u.c d˜a cho ta c´o thˆe’
biˆe´n dˆo’i so. bˆo. biˆe’u th´u.c d´o sao cho qu´a tr`ınh t´ınh da.o h`am do.n gia’ n
ho.n. Ch˘a’ ng ha. n nˆe´u biˆe’u th´u.c d´o l`a logarit th`ı c´o thˆe’ su.’ du. ng c´ac
t´ınh chˆa´t cu’ a logarit dˆe’ biˆe´n dˆo’i... rˆo` i t´ınh da. o h`am. Trong nhiˆe`u
tru.`o.ng ho.
.p khi t´ınh da.o h`am ta nˆen lˆa´y logarit h`am d˜a cho rˆo` i ´ap
du. ng cˆong th´u.c da. o h`am loga · lny(x) = d
dx y0(x)
y(x) .c cho bo.’ i phu.o.ng tr`ınh 2) Nˆe´u h`am kha’ vi trˆen mˆo. t khoa’ ng du.o. F (x, y) = 0. F (x, y) = 0 th`ı da. o h`am y0(x) c´o thˆe’ t`ım t`u. phu.o.ng tr`ınh
d
dx C ´AC V´I DU. 65 8.1. D- a. o h`am V´ı du. 1. T´ınh da.o h`am y0 nˆe´u: ; x 6= π(2n + 1), n ∈ N 1) y = ln 3r ex 1 + cos x 3 √ 2) y = , x 6= πn, n ∈ N. 1 + x2
x4 sin7 x Gia’ i. 1) Tru.´o.c hˆe´t ta do.n gia’ n biˆe’u th´u.c cu’ a h`am y b˘a`ng c´ach
.a v`ao c´ac t´ınh chˆa´t cu’ a logarit. Ta c´o du. − y = lnex − ln(1 + cos x) = ln(1 + cos x). 1
3 1
3 x
3 1
3 Do d´o 1 + tg x
2 · − = + = y0 = (cos x)0
1 + cos x 1
3 1
3 sin x
1 + cosx 3 2) O .i ho.n ca’ l`a x´et h`am z = ln|y|. Ta c´o · ⇒ · = = = y (*) 1
1
3
3
.’ dˆay tiˆe.n lo.
dz
dx dy
dx 1
y dy
dx dy
dx dz
dx dz
dy
Viˆe´t h`am z du.´o.i da. ng ln|x| − 7ln| sin x| x = ln|y| = ln(1 + x2) − 4
3 ⇒ · − = − 7 dz
dx 2x
1 + x2 cos x
sin x 4
3x
.c v`ao (∗) ta c´o Thˆe´ biˆe’u th´u.c v`u.a thu du.o. 3 (cid:16) √ − (cid:17). N = − 7 dy
dx 2x
1 + x2 4
3x cos x
sin x 1 + x2
x4 sin7 x V´ı du. 2. T´ınh da.o h`am y0 nˆe´u: 1) y = (2 +cos x)x, x ∈ R; 2) y = x2x,
x > 0. Gia’ i. 1) Theo di.nh ngh˜ıa ta c´o y = exln(2+cos x). 66 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 0
y0 = exln(2+cos x)(cid:2)xln(2 + cos x)(cid:3) T`u. d´o i, x ∈ R. = exln(2+cos x)hln(2 + cos x) − x sin x
2 + cos x y0 = e2xlnx[2xlnx]0 = e2xlnxh 2x + 2xln2 · lnxi + ln2 · lnx(cid:17). N 2) V`ı y = e2xlnx nˆen v´o.i x > 0 ta c´o
1
x
1
= 2xx2x(cid:16)
x .c v´o.i h`am y = x + x5, V´ı du. 3. T´ınh da. o h`am cˆa´p 2 cu’ a h`am ngu.o.
x ∈ R. Gia’ i. H`am d˜a cho liˆen tu. c v`a do.n diˆe.u kh˘a´p no.i, da. o h`am y0 = 1 + 5x4 khˆong triˆe.t tiˆeu ta. i bˆa´t c´u. diˆe’m n`ao. Do d´o 0 · = x0
y = 1
1 + 5x4 1
y0
x y = x (cid:17) · x0 · N x00
yy = (cid:16) Lˆa´y da. o h`am d˘a’ ng th´u.c n`ay theo y ta thu du.o.
.c
−20x3
(1 + 5x4)3 1
1 + 5x4 xx. V´ı du. 4. Gia’ su.’ h`am y = f (x) du.o.
.c cho du.´o.i da. ng tham sˆo´ bo.’ i c´ac
cˆong th´u.c x = x(t), y = y(t), t ∈ (a; b) v`a gia’ su.’ x(t), y(t) kha’ vi cˆa´p
2 v`a x0(t) 6= 0 t ∈ (a, b). T`ım y00 Gia’ i. Ta c´o x = · = = ⇒ y0 dy
dx y0
t
x0
t y0
t
x0
t dy
dt
dx
dt 0 Lˆa´y da. o h`am hai vˆe´ cu’ a d˘a’ ng th´u.c n`ay ta c´o t 0
(cid:17)
t (cid:17) · · t0 y0
t
x0
t 1
x0
t x = (cid:16)
tx00 tt 3 tt − y0
x0
t y0
y00
xx = (cid:16)
t
x0
t
ty00
x0 = · N 67 8.1. D- a. o h`am V´ı du. 5. Gia’ su.’ y = y(x), |x| > a l`a h`am gi´a tri. du.o.ng cho du.´o.i
da. ng ˆa’n bo.’ i phu.o.ng tr`ınh − = 1. x2
a2 y2
b2 xx. T´ınh y00 Gia’ i. Dˆe’ t`ım y0 ta ´ap du. ng cˆong th´u.c F (x, y) = 0. d
dx
.p n`ay ta c´o Trong tru.`o.ng ho. (cid:16) − − 1(cid:17) = 0. d
dx x2
a2 y2
b2 Lˆa´y da. o h`am ta c´o − (8.1) y0
x = 0, 2x
a2 x = , |x| > 0, y > 0. (8.2) ⇒y0 2y
b2
b2x
a2y 2 − .c
Lˆa´y da. o h`am (8.1) theo x ta thu du.o. − y00
xx = 0 1
b2 (cid:0)y0
x(cid:1) y
b2 1
a2
.c y00
x: 2i = i h h − y00
xx = − (cid:0)y0
x(cid:1) 1
y 1
y b2
a2 b4
a4 x2
y2 h − i = − = − , y > 0. N v`a t`u. (8.2) ta thu du.o.
b2
a2
b4
a2y3 x2
a2 y2
b2 b4
a2y3 ; 2) y = x2 cos 2x. V´ı du. 6. T´ınh y(n) nˆe´u: 1) y = 1
x2 − 4
Gia’ i. 1) Biˆe’u diˆe˜n h`am d˜a cho du.´o.i da. ng tˆo’ng c´ac phˆan th´u.c co. ba’ n h i − = 1
x2 − 4 1
4 1
x − 2 1
x + 2 68 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n (n) (n) (n) v`a khi d´o (cid:17) h(cid:16) (cid:17) (cid:17) (cid:16) − (cid:16) i. = 1
x2 − 4 1
4 1
x − 2 1
x + 2 (n) Do (cid:16) (cid:17) = (−1)(−2) · · · (−1 − n + 1)(x ± 2)−1−n 1
x ± 2 = (−1)nn! 1
(x ± 2)n+1 (n) nˆen (cid:17) h (cid:16) − i. = 1
x2 − 4 (−1)nn!
4 1
(x − 2)n+1 1
(x + 2)n+1 2) Ta ´ap du. ng cˆong th´u.c Leibniz dˆo´i v´o.i da. o h`am cu’ a t´ıch n(x2)0(cos 2x)n−1 (x2 cos 2x) = C 0 nx2(cos 2x)(n) + C 1
n(x2)0(cos 2x)n−2. + C 2 (k) = 0 ∀ k > 2. C´ac sˆo´ ha. ng c`on la. i dˆe`u = 0 v`ı (cid:0)x2(cid:1) ´Ap du. ng cˆong th´u.c (cid:17) (cos 2x)(n) = 2n cos (cid:16)2x + nπ
2 .c
ta thu du.o. (cid:17) (x2 cos 2x)(n) = 2n(cid:16)x2 − n(n − 1)
4 nπ
2 (cid:17). N + 2nnx sin (cid:16)2x + (cid:17) cos (cid:16)2x +
nπ
2
V´ı du. 7. V´o.i gi´a tri. n`ao cu’ a a v`a b th`ı h`am x 6 0, ex,
f (x) = x2 + ax + b, x > 0 69 8.1. D- a. o h`am c´o da. o h`am trˆen to`an tru. c sˆo´. Gia’ i. R˜o r`ang l`a h`am f (x) c´o da. o h`am ∀ x > 0 v`a ∀ x < 0. Ta chı’ cˆa` n x´et diˆe’m x0 = 0. V`ı h`am f (x) pha’ i liˆen tu. c ta. i diˆe’m x0 = 0 nˆen f (x) lim
x→0+0 f (x) = lim
x→0−0 f (x) = lim
x→0 t´u.c l`a (x2 + ax + b) = b = e0 = 1 ⇒ b = 1. lim
x→0+0 −(0) = ex(cid:12) +(0) = (x0 + ax + b)0(cid:12) Tiˆe´p d´o, f 0 = a v`a f 0 (cid:12)x0=0 (cid:12)x0=0 = 1.
Do d´o f 0(0) tˆo` n ta. i nˆe´u a = 1 v`a b = 1. Nhu. vˆa. y v´o.i a = 1, b = 1
h`am d˜a cho c´o da. o h`am ∀ x ∈ R. N B `AI T ˆA. P T´ınh da. o h`am y0 cu’ a h`am y = f (x) nˆe´u: − − x3 + + 2. (DS. + ) √
1. y = 4 5
x2 3
x3 3
√
4 4 x 10
x3 9
x4 x ) (DS. 2. y = log2x + 3log3x. ln24
xln2 · ln3 (cid:17) (DS. 5xln5 + 6xln6 − 7−xln7) . 3. y = 5x + 6x + (cid:16) 1
7
√ √ x2 + 2x + 3). (DS. 4. y = ln(x + 1 + ) 1
x2 + 2x + 3 ) 5. y = tg5x. (DS. 10
sin 10x √ 6. y = ln(ln ) x). (DS. 1
√
2xln x . (DS. ) 7. y = ln r1 + 2x
1 − 2x 2
1 − 4x2 70 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n √ √ √ . 8. y = xarctg 2x − 1 − (DS. arctg 2x − 1) 2x − 1
2
(DS. 3x2 sin 2x3) 9. y = sin2 x3. √ 10. y = sin4 x + cos4 x. x. x(1 +
√
x
2 √ (DS. − sin 4x)
√
√
e x) xe (DS. 11. y = ) 1
cos x ) 12. y = e 1
cos x . (DS. e sin x
cos2 x ) 13. y = e 1
lnx . (DS. 1
lnx
−e
xln2x √ √ ) e4x + 1. (DS. 14. y = ln(cid:0)e2x + 2e2x
e4x + 1 r e4x 15. y = ln . ) (DS. e4x + 1 − ) (DS. − 16. y = log5 cos 7x. 2
e4x + 1
7tg7x
ln5 √ √ ) 1 + x. (DS. − 17. y = log7 cos tg
√
2 1 + x
1 + xln7 −
18. y = arccos(cid:0)e x2
2 √ (DS. ) x2
2 (cid:1). 19. y = tg sin cos x. ) (DS. xe
1 − e−x2
− sin cos(cos x)
cos2(sin cos x) √ (sin 6x − 3x)) 20. y = ex2cotg3x. (DS. 1+lnx. x ) (DS. 21. y = e xec2cotg3x
sin2 3x
√
1+lnx
e
√
2x 1 + lnx −2(1 − lnx)) 22. y = x 1
x . (DS. x 1 (DS. xx(1 + lnx)) 23. y = ex. 71 8.1. D- a. o h`am 24. y = xsin x. i) 25. y = (tgx)sin x. (DS. (tgx)sin xh cos xlntgx + (DS. xsin x cos x · lnx + xsin x−1 sin x)
1
cos x + lnx · cos xi) 26. y = xsin x. (DS. xsin xh sin x
x 27. y = xx2. (DS. xx2+1(1 + 2lnx)) x + lnx)) 28. y = xex . (DS. exxex (cid:16) 1 ) (DS. − 29. y = logx7. . (DS. ) 30. y = 1
2a x − a
x + a 1
xlnxlog7x
1
x2 − a2 ln(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12) ) 31. y = sin ln|x|. (DS. cos ln|x|
x (DS. cotgx) 32. y = ln| sin x|.
√ √ 33. y = ln|x + ). x2 + 1|. (DS. 1
x2 + 1 .c
Trong c´ac b`ai to´an sau dˆay (34-40) t´ınh da. o h`am cu’ a h`am y du.o. cho du.´o.i da. ng tham sˆo´. xx ? 34. x = a cos t, a sin t, t ∈ (0, π). y00 ) (DS. − 1
a sin3 t xx ? (DS. − ) 35. x = t3, y = t2. y00 2
9t4 xx ? 36. x = 1 + eat, y = at + e−at. y00 xx ? ) 37. x = a cos3 t, y = a sin3 t. y00 (DS. xx ? (DS. ) 38. x = et cos t, y = et sin t. y00 xx ? (DS. − 39. x = t − sin t, y = 1 − cos t. y00 ) (DS. 2e−3at − e−2at)
1
3a sin t cos4 t
2
et(cos t − sin t)3
1
4 sin4 t
2 xx ? (DS. ). 40. x = t2 + 2t, y = ln(1 + t). y00 −1
4(1 + t)4 72 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n Trong c´ac b`ai to´an sau dˆay (41-47) t´ınh da. o h`am y0 ho˘a. c y00 cu’ a h`am ˆa’n du.o.
√ 41. x + ) .c x´ac di.nh bo.’ i c´ac phu.o.ng tr`ınh d˜a cho
xy + y = a. y0 ? (DS. 42. arctg ) (DS. = lnpx2 + y2. y0 ? 2a − 2x − y
x + 2y − a
x + y
x − y y
x ) 43. ex sin y − e−y cos x = 0. y0 ? (DS. − ex sin y + e−y sin x
ex cos y + e−y cos x ) (cid:17) = 0. y0 ? (DS. 44. x2y + arctg(cid:16) y
x (DS. ) 45. ex − ey = y − x. y00 ? −2x3y − 2xy3 + y
x4 + x2y2 + x
(ey − ex)(ex+y − 1)
(ey + 1)3 46. x + y = ex−y. y00 ? (DS. ) 47. y = x + arctgy. y00 ? (DS. ). 4(x + y)
(x + y + 1)3
−(2y2 + 2)
y5 .c
Trong c´ac b`ai to´an sau dˆay (48-52) t´ınh da.o h`am cu’ a h`am ngu.o. v´o.i h`am d˜a cho. y ? y = 48. y = x + x3, x ∈ R. x0 (DS. x0 ) y ? y = , y > 0) (DS. x0 49. y = x + lnx, x > 0. x0 1
1 + 3x2
x
x + 1 y ? y = , y ∈ R) 50. y = x + ex. x0 (DS. x0 y ? y = ) (DS. x0 51. y = chx, x > 0. x0 y ? y = 52. y = , x < 0. x0 (DS. x0 , y ∈ (0, 1)). x2
1 + x2 1
1 + y − x
1
py2 − 1
x3
2y2 53. V´o.i gi´a tri. n`ao cu’ a a v`a b th`ı h`am x3
f (x) = nˆe´u x 6 x0,
ax + b nˆe´u x > x0 73 8.1. D- a. o h`am 0, b = −2x3 0). liˆen tu. c v`a kha’ vi ta. i diˆe’m x = x0 ? (DS. a = 3x2 54. X´ac di.nh α v`a β dˆe’ c´ac h`am sau: a) liˆen tu. c kh˘a´p no.i; b) kha’ vi
kh˘a´p no.i nˆe´u αx + β nˆe´u x 6 1
1) f (x) = x2 nˆe´u x > 1 nˆe´u |x| < 1,
2) f (x) = nˆe´u |x| > 1. α + βx2
1
|x| (DS. 1) a) α + β = 1, b) α = 2, β = −1; 2) a) α + β = 1, b) ). , β = − 3
2 1
2 α =
55. Gia’ su.’ h`am y = f (x) x´ac di.nh trˆen tia (−∞, x0) v`a kha’ vi bˆen
tr´ai ta. i diˆe’m x = x0. V´o.i gi´a tri. n`ao cu’ a a v`a b th`ı h`am f (x)
f (x) = ax2 + b nˆe´u x 6 x0,
nˆe´u x > x0 kha’ vi ta. i diˆe’m x = x0 (x0 6= 0) ? (DS. a = , b = f (x0) − f 0(x0 − 0)). x0
2 f 0(x0 − 0)
2x0 Trong c´ac b`ai to´an (56-62) t´ınh da. o h`am y00 nˆe´u (DS. 2e−x2 (2x2 − 1)) 56. y = e−x2. ) 57. y = tgx. (DS. 2 sin x
cos3 x √ 1 + x2. (DS. 58. y = . (DS. 59. y = arcsin 1
(1 + x2)3/2 )
x
(4 − x2)3/2 ) . (DS. ) 60. y = arctg x
2
1
x 2x
(1 + x2)2 74 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 2 − x2
√ 61. y = x arcsinx. (DS. ) 1 − x2 (1 − x2)
(DS. exf 0(ex) + e2xf 00(ex)). 62. y = f (ex). . (DS. ) 63. y = arctg x
2 Trong c´ac b`ai to´an (63-69) t´ınh da. o h`am cˆa´p 3 cu’ a y nˆe´u:
4(3x − 4)
(4 + x2)3
(DS. e−x(3 − x)) 64. y = xe−x. 65. y = ex cos x. (DS. −2ex(cos x + sin x)) (DS. −2ex(cos x + sin x)) 66. y = x2 sin x. 67. y = x32x. (DS. 2x(x3ln32 + 9x2ln2x + 18xln2 + 6)) 68. y = x2 sin 2x. (DS. −4(2x2 cos 2x + 6x sin 2x − 3 cos 2x)) 69. y = (f (x2). (DS. 12xf 00(x2) + 8x3f 000(x2)). n 70. y = sin 3x. (cid:17)) Trong c´ac b`ai to´an (70-84) t´ınh da. o h`am y(n) nˆe´u
(DS. 3n sin (cid:16)3x + nπ
2 2 (cid:16) (cid:17) (DS. e x ) 72. y = e x
2 . 73. y = 23x. 1
2
(DS. 23x(3ln2)n) (cid:17)) 74. y = cos2 x. (DS. 2n−1 cos (cid:16)2x + n · π
2 (DS. 4nn!) 75. y = (4x + 1)n. ) 76. y = ln(ax + b). (DS. (−1)n−1(n − 1)! (cid:17)) (DS. 4n−1 cos (cid:16)4x + 77. y = sin4 x + cos4 x. + cos 4x. Chı’ dˆa˜n. Ch´u.ng minh r˘a`ng sin4 x + cos4 x = an
(ax + b)n
nπ
2
3
4 1
4 (cid:17) − sin (cid:16)x + sin (cid:16)3x + n · (cid:17)) 78. y = sin3 x. (DS. 3
4 nπ
2 3n
4 π
2 Chı’ dˆa˜n. D`ung cˆong th´u.c sin 3x = 3 sin x − 4 sin3 x. 75 8.2. Vi phˆan 79. y = sin αx sin βx. ] − ]) (DS. (α − β)n cos[(α − β)x + n (α + β)n cos[(α + β)x + n 1
2 π
2 1
2 π
2 Chı’ dˆa˜n. Biˆe´n dˆo’i t´ıch th`anh tˆo’ng. 80. y = eαx sin βx. (DS. eαxn sin βxhαn − αn−2β2 + . . . i+ + cos βxhnαn−1β − αn−3β3 + . . . i) n(n − 1)
1 · 2
n(n − 1)(n − 2)
1 · 2 · 3 Chı’ dˆa˜n. D`ung quy t˘a´c Leibniz. (DS. ex[3x2 + 6nx + 3n(n − 1) − 4]) 81. y = ex(3x2 − 4). (cid:16) 82. y = ln > 0(cid:17) ax + b
ax − b ax + b
ax − b − i) (DS. (−1)n−1an(n − 1)!h 1
(ax + b)n 1
ax − b)n h . (DS. i) 83. y = x
x2 − 4x − 12 (−1)nn!
4 1
(x − 2)n+1 . (DS. (−1)nn!h i) 84. y = 3 − 2x2
2x2 + 3x − 2 3
(x − 6)n+1 +
2n
(2x − 1)n+1 + 1
(x + 2)n+1 Chı’ dˆa˜n. Dˆe’ gia’ i b`ai 83 v`a 84 cˆa` n biˆe’u diˆe˜n h`am d˜a cho du.´o.i da. ng tˆo’ng c´ac phˆan th´u.c do.n gia’ n. 8.2.1 Vi phˆan cˆa´p 1 Gia’ su.’ h`am y = f (x) x´ac di.nh trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m x0 v`a
∆x = x − x0 l`a sˆo´ gia cu’ a biˆe´n dˆo. c lˆa. p. H`am y = f (x) c´o vi phˆan cˆa´p
1 (vi phˆan th´u. nhˆa´t) ta. i diˆe’m x0 nˆe´u khi dˆo´i sˆo´ di.ch chuyˆe’n t`u. gi´a tri.
x = x0 dˆe´n gi´a tri. x = x0 + ∆x sˆo´ gia tu.o.ng ´u.ng cu’ a h`am f (x) c´o thˆe’ 76 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n biˆe’u diˆe˜n du.´o.i da. ng (8.3) ∆f (x0) ≡ f (x0 + ∆x) − f (x0) = D(x0)∆x + o(∆x) o(∆x)
∆x → 0 khi ∆x → 0. T´ıch
.c
.c go. i l`a vi phˆan cˆa´p 1 cu’ a h`am f (x) ta. i diˆe’m x0 v`a du.o. trong d´o D(x0) khˆong phu. thuˆo. c ∆x v`a
D(x0)∆x du.o.
k´y hiˆe.u dx. dy ≡ df ≡ Sˆo´ gia ∆x cu’ a biˆe´n dˆo. c lˆa. p x du.o. dy
dx
.c go. i l`a vi phˆan cu’ a biˆe´n dˆo. c lˆa. p, t´u.c l`a theo di.nh ngh˜ıa: dx = ∆x.
D- i.nh l´y 8.2.1. H`am y = f (x) c´o vi phˆan cˆa´p 1 ta. i diˆe’m x0 khi v`a
chı’ khi h`am d´o c´o da. o h`am h˜u.u ha. n ta. i d´o v`a D(x0) = f 0(x0). Vi phˆan df (x0) cu’ a h`am f ta. i diˆe’m x0 biˆe’u diˆe˜n qua da. o h`am f 0(x0) bo.’ i cˆong th´u.c df (x0) = f 0(x0)dx
(8.4)
Cˆong th´u.c (8.4) cho ph´ep t´ınh vi phˆan cu’ a c´ac h`am, nˆe´u biˆe´t da. o h`am
cu’ a ch´ung. T`u. (8.3) suy ra dx → 0. y(x0 + ∆x) = y(x0) + df (x0) + o(dx), Nˆe´u df (x0) 6= 0 th`ı dˆe’ t´ınh gi´a tri. gˆa` n d´ung cu’ a h`am f (x) ta. i diˆe’m
x0 + ∆x ta c´o thˆe’ ´ap du. ng cˆong th´u.c (8.5) y(x0 + ∆x) ≈ y(x0) + df (x0) Vi phˆan cˆa´p 1 c´o c´ac t´ınh chˆa´t sau.
1+ d(αu + βv) = αdu + βdv, d(cid:16) (cid:17) = , v 6= 0. d(uv) = udv + vdu,
vdu − udv
v2 u
v 8.2. Vi phˆan 77 8.2.2 Vi phˆan cˆa´p cao 2+ Cˆong th´u.c vi phˆan dy = f 0(x)dx luˆon luˆon tho’a m˜an bˆa´t luˆa. n
x l`a biˆe´n dˆo. c lˆa. p hay l`a h`am cu’ a biˆe´n dˆo. c lˆa. p kh´ac. T´ınh chˆa´t n`ay
du.o. .c go. i l`a t´ınh bˆa´t biˆe´n vˆe` da. ng cu’ a vi phˆan cˆa´p 1. Gia’ su.’ x l`a biˆe´n dˆo.c lˆa. p v`a h`am y = f (x) kha’ vi trong lˆan cˆa. n n`ao
d´o cu’ a diˆe’m x0. Vi phˆan th´u. nhˆa´t df = f 0(x)dx l`a h`am cu’ a hai biˆe´n
x v`a dx, trong d´o dx l`a sˆo´ t`uy ´y khˆong phu. thuˆo. c v`ao x v`a do d´o (dx)0 = 0. Vi phˆan cˆa´p hai (hay vi phˆan th´u. hai) d2f cu’ a h`am f (x) ta. i diˆe’m
.c di.nh ngh˜ıa nhu. l`a vi phˆan cu’ a h`am df = f 0(x)dx ta. i diˆe’m x0 x0 du.o.
v´o.i c´ac diˆe`u kiˆe.n sau dˆay: 1) df pha’ i du.o. .c xem l`a h˘a`ng sˆo´); .c xem l`a h`am cu’ a chı’ mˆo. t biˆe´n dˆo. c lˆa. p x (n´oi c´ach
kh´ac: khi t´ınh vi phˆan cu’ a f 0(x)dx ta cˆa` n t´ınh vi phˆan cu’ a f 0(x), c`on
dx du.o. du.o. 2) Sˆo´ gia cu’ a biˆe´n dˆo. c lˆa.p x xuˆa´t hiˆe.n khi t´ınh vi phˆan cu’ a f 0(x)
.c xem l`a b˘a`ng sˆo´ gia dˆa` u tiˆen, t´u.c l`a b˘a`ng dx.
Nhu. vˆa. y theo di.nh ngh˜ıa ta c´o d2f = d(df ) = d(f 0(x)dx) = (df 0(x))dx = f 00(x)dxdx = f 00(x)(dx)2 hay l`a d2f = f 00(x)dx2 , dx2 = (dx)2. (8.6) .c
B˘a`ng phu.o.ng ph´ap quy na. p, dˆo´i v´o.i vi phˆan cˆa´p n ta thu du.o. cˆong th´u.c dnf = f (n)(x)dxn (8.7) 78 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n .c xem l`a b˘a`ng 0, t´u.c Vi phˆan cˆa´p n (n > 1) cu’ a biˆe´n dˆo.c lˆa. p x du.o. l`a dnx = 0 v´o.i n > 1. (8.8) Nˆe´u ∃ dnf v`a ∃ dng v`a α, β ∈ R th`ı (8.9) dn(αf + βg) = αdnf + βdng ndn−kf · dkg. n
X
k=0 C k (8.10) dnfg = Ch´u ´y. 1) Khi n > 1, c´ac cˆong th´u.c (8.6) v`a (8.7) chı’ d´ung khi x
.c
.p y = y(x(t)) cˆong th´u.c (8.6) du.o. l`a biˆe´n dˆo. c lˆa. p. Dˆo´i v´o.i h`am ho.
kh´ai qu´at nhu. sau: xdx) = d(y0 x)dx + y0 xd(dx) d2y = d(dy) = d(y0 v`a do d´o xxdx2 + y0 xd2x. (8.11) d2y = y00 .p khi x l`a biˆe´n dˆo. c lˆa. p th`ı d2x = 0 (xem (8.8)) v`a Trong tru.`o.ng ho.
cˆong th´u.c (8.11) tr`ung v´o.i (8.6). 2) Khi t´ınh vi phˆan cˆa´p n ta c´o thˆe’ biˆe´n dˆo’i so. bˆo. h`am d˜a cho.
Ch˘a’ ng ha. n nˆe´u f (x) l`a h`am h˜u.u ty’ th`ı cˆa` n khai triˆe’n n´o th`anh tˆo’ng
.ng gi´ac
h˜u.u ha. n c´ac phˆan th´u.c h˜u.u ty’ co. ba’ n; nˆe´u f (x) l`a h`am lu.o.
th`ı cˆa` n ha. bˆa. c nh`o. c´ac cˆong th´u.c ha. bˆa. c,...
3) T`u. cˆong th´u.c (8.7) suy ra r˘a`ng f (n)(x) = dnf
dxn t´u.c l`a da. o h`am cˆa´p n cu’ a h`am y = f (x) ta. i mˆo. t diˆe’m n`ao d´o b˘a`ng ty’
sˆo´ gi˜u.a vi phˆan cˆa´p n cu’ a h`am f (x) chia cho l˜uy th`u.a bˆa. c n cu’ a vi
phˆan cu’ a dˆo´i sˆo´. 8.2. Vi phˆan 79 C ´AC V´I DU. √ V´ı du. 1. T´ınh vi phˆan df nˆe´u
1) f (x) = ln(arctg(sin x)); 2) f (x) = x . 64 − x2 +64arcsin x
8 Gia’ i. 1) ´Ap du. ng c´ac t´ınh chˆa´t cu’ a vi phˆan ta c´o = df = d[arctg(sin x)]
arctg(sin x) d(sin x)
(1 + sin2 x)arctg(sin x) · = cos xdx
(1 + sin2 x)arctg(sin x) 2) √ i df = d[x 64 − x2] + dh64arcsin
√ √ (cid:17) 64 − x2 + = xd x
8
(cid:17) d(cid:16) √ x
8
64 − x2dx + 64d(cid:16)arcsin
x
8 64 − x2dx + 64 · + = x d(64 − x2)
√
64 − x2
2 r 1 − x2
64 √ √ + 64 − x2dx + 64 = dx
64 − x2 −x2dx
√
64 − x2
√
64 − x2dx, = 2 |x| < 8. N V´ı du. 2. T´ınh vi phˆan cˆa´p 2 cu’ a c´ac h`am
1) f (x) = xe−x, nˆe´u x l`a biˆe´n dˆo. c lˆa. p;
2) f (x) = sin x2 nˆe´u a) x l`a biˆe´n dˆo. c lˆa.p,
b) x l`a h`am cu’ a mˆo. t biˆe´n dˆo. c lˆa. p n`ao d´o. Gia’ i. 1) Phu.o.ng ph´ap I. Theo di.nh ngh˜ıa vi phˆan cˆa´p 2 ta c´o d2f = d[df ] = d[xde−x + e−xdx] = d(−xe−xdx + e−xdx) = −d(xe−x)dx + d(e−x)dx
= −(xde−x + e−xdx)dx − e−xdx2
= xe−xdx2 − e−xdx2 − e−xdx2 = (x − 2)e−xdx2. 80 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n Phu.o.ng ph´ap II. T´ınh da. o h`am cˆa´p hai f 00(x) ta c´o f 00(x) = (xe−x)00 = (e−x − xe−x)0 = −e−x − e−x + xe−x = (x − 2)e−x v`a theo cˆong th´u.c (8.6) ta c´o d2f = (x − 2)e−xdx2. 2) a) Phu.o.ng ph´ap I. Theo di.nh ngh˜ıa vi phˆan cˆa´p hai ta c´o d2f = d[d sin x2] = d[2x cos x2dx] = d[2x cos x2]dx xx ta c´o = (cid:0)2 cos x2dx + 2x(− sin x2)2xdx(cid:1)dx
= (2 cos x2 − 4x2 sin x2)dx2. Phu.o.ng ph´ap II. T´ınh da. o h`am cˆa´p hai f 00 f 0
x = 2x cos x2, f 00
xx = 2 cos x2 − 4x2 sin x2 .c
v`a theo (8.6) ta thu du.o. d2f = (2 cos x2 − 4x2 sin x2)dx2. b) Nˆe´u x l`a biˆe´n trung gian th`ı n´oi chung d2x 6= 0 v`a do d´o ta c´o d2f = d(2x cos x2dx) = (2x cos x2)d2x + [d(2x cos x2)]dx
= 2x cos x2d2x + (2 cos x2 − 4x2 sin x2)dx2. N V´ı du. 3. ´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung c´ac gi´a tri.: 5r 2 − 0, 15
2 + 0, 15 ; 2) arcsin 0, 51; 3) sin 29◦. 1) Gia’ i. Cˆong th´u.c co. ba’ n dˆe’ ´u.ng du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung l`a ∆f (x0) ≈ df (x0) ⇒ f (x0 + ∆x) − f (x0) ≈ f 0(x0)∆x
⇒ f (x0 + ∆x) ≈ f (x0) + f 0(x0)∆x 8.2. Vi phˆan 81 .c hiˆe.n nhu. sau: T`u. d´o, dˆe’ t´ınh gˆa` n d´ung c´ac gi´a tri. ta cˆa` n thu.
1+ Chı’ ra biˆe’u th´u.c gia’ i t´ıch dˆo´i v´o.i h`am m`a gi´a tri. gˆa` n d´ung cu’ a n´o cˆa` n pha’ i t´ınh. 2+ Cho. n diˆe’m M0(x0) sao cho gi´a tri. cu’ a h`am v`a cu’ a da. o h`am cˆa´p 1 cu’ a n´o ta. i diˆe’m ˆa´y c´o thˆe’ t´ınh m`a khˆong d`ung ba’ ng. 3+ Tiˆe´p dˆe´n l`a ´ap du. ng cˆong th´u.c v`u.a nˆeu.
1) T´ınh gˆa` n d´ung 5r 2 − 0, 15
2 + 0, 15
Sˆo´ d˜a cho l`a gi´a tri. cu’ a h`am y = 5r2 − x
2 + x ta. i diˆe’m x = 0, 15. Ta d˘a. t x0 = 0; ∆x = 0, 15. Ta c´o · = − y0 = ⇒ y0(x0) = y0(0) = − −4 5r 2 − x
2 + x
5(4 − x2) 4y
5(4 − x2) 1
5 Do d´o v`ı y(0) = 1 nˆen y(0, 15) ≈ y(0) + y0(0)∆x · (0, 15) = 1 − 0, 03 = 0, 97. = 1 − 1
5 2) T´ınh gˆa` n d´ung arcsin 0, 51.
X´et h`am y = arcsin x. Sˆo´ cˆa` n t´ınh l`a gi´a tri. cu’ a h`am ta. i diˆe’m 0, 51; t´u.c l`a y(0, 51). D˘a. t x0 = 0, 5; ∆x = 0, 01. Khi d´o ta c´o ∆x arcsin(x0 + ∆x ≈ arcsinx0 + (arcsinx)0 x=x0
⇒ arcsin(0, 5 + 0, 01) ≈ arcsin0, 5 + (arcsinx)0(cid:12) · 0, 01 = + (cid:12)x=0,5
× (0, 01). π
6 1
p1 − (0, 5)2 82 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n √ 0, 75 ≈ 0, 88 v`a do d´o C´o thˆe’ t´ınh gˆa` n d´ung p1 − (0, 5)2 = + 0, 011 ≈ 0, 513. arcsin0, 51 ≈ π
6 3) Sˆo´ sin 29◦ l`a gi´a tri. cu’ a h`am y = sin x khi x = π
180 · × 30 = ; y(cid:16) , y = cos x ⇒ y0(cid:16) = (cid:17) = (cid:17) = cos x0 = π
180 π
6 π
6 1
2 × 29. Ta d˘a. t
√
3
2 π
6 π
6 − = − . Do d´o D˘a. t ∆x = x − x0 = 29π
180 π
6 π
180 √ (cid:16) − (cid:17) + y0(cid:16) (cid:17) · ∆x = + (cid:17) ≈ 0, 48. N sin 29◦ ≈ y(cid:16) π
6 π
6 1
2 3
2 π
180 B `AI T ˆA. P T´ınh vi phˆan df nˆe´u: 1. f (x) = arctg . (DS. df = 1
x ) −dx
1 + x2 )
(DS. 2tg2xln2 · 2tgx · 2. f (x) = 2tg2x. dx
cos2 x 3. f (x) = arccos(2x). (DS. − ) 2xln2dx
√
1 − e2x
(DS. x2(1 + 3lnx)dx) 4. f (x) = x3lnx.
√ √ √ ) 5. f (x) = cos2( x). (DS. −2 cos x · sin x · dx
√
x
2 6. f (x) = (1 + x2)arcotgx. (DS. (2xarccotgx − 1)dx) 7. f (x) = . (DS. dx) 1 − xarctgx
(1 + x2)3/2 (DS. 3 sin 2x sin 4xdx) √ arctgx
√
1 + x2
8. f (x) = sin3 2x.
√ x dx) x). (DS. 9. f (x) = ln(sin cotg
√
2 x cos x 8.2. Vi phˆan 83 cos x . dx) 10. f (x) = e− 1 (DS. −tgx · e− 1
cos x 11. f (x) = 2−x2 . √ 12. f (x) = arctg x2 + 1. (DS. ) √ √ √ √ (cid:16)arctg 13. f (x) = (cid:17)dx) xarctg (DS. x. x + x
1 + x (DS. −2xe−x2 ln2dx)
2xdx
2 + x2
1
√
2 x i √ xh2arcsinx − x
1 − x2 14. f (x) = . (DS. dx). x2
arcsinx (arcsinx)2 T´ınh vi phˆan cˆa´p tu.o.ng ´u.ng cu’ a c´ac h`am sau (DS. 4−x2 2ln4(2x2ln4 − 1)(dx)2) 15. f (x) = 4−x2 ; d2f ? (DS. (dx)2) 16. f (x) = pln2x − 4. d2f ? 4lnx − 4 − ln3x
x2p(lnx − 4)3 (DS. −4 sin 2x(dx)3) 17. f (x) = sin2 x. d3f ?
√ 18. f (x) = (DS. (dx)4) x − 1, d4f ? (DS. − 19. f (x) = xlnx, d5f ? −15
16(x − 1)7/2
6
x4 (dx)5, x > 0)
(DS. (10 cos x − x sin x)(dx)10) 20. f (x) = x sin x; d10f ? Su.’ du. ng cˆong th´u.c gˆa` n d´ung ∆f ≈ df (khi f 0(x) 6= 0) dˆe’ t´ınh gˆa` n d´ung c´ac gi´a tri. sau √ 3, 98. (DS. 1,955) 21. y = 26, 19. (DS. 2,97) √
22. y = 3 . (DS. 0,35) 23. y = s (2, 037)2 − 3
(2, 037)2 + 5 (DS. 0,85) 24. y = cos 31◦. 84 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 25. y = tg45◦100. (DS. 0,99) 26. y = ln(10, 21). (DS. 1,009) 27. y = sin 31◦. (DS. 0,51) 28. y = arcsin0, 54. (DS. 0,57) 29. y = arctg(1, 05). (DS. 0,81) 30. y = (1, 03)5. (DS. 1,15) 8.3.1 C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi
D- i.nh l´y Rˆon (Rolle). Gia’ su.’ : i) f (x) liˆen tu. c trˆen doa. n [a, b].
ii) f (x) c´o da. o h`am h˜u.u ha. n trong (a, b).
iii) f (a) = f (b). Khi d´o tˆo` n ta. i diˆe’m ξ : a < ξ < b sao cho f (ξ) = 0.
D- i.nh l´y Lagr˘ang (Lagrange). Gia’ su.’ :
i) f (x) liˆen tu. c trˆen doa. n [a, b].
ii) f (x) c´o da. o h`am h˜u.u ha. n trong (a, b). Khi d´o t`ım du.o. .c ´ıt nhˆa´t mˆo. t diˆe’m ξ ∈ (a, b) sao cho = f 0(ξ) (8.12) f (b) − f (a)
b − a hay l`a f (b) = f (a) + f 0(ξ)(b − a). (8.13) Cˆong th´u.c (8.12) go. i l`a cˆong th´u.c sˆo´ gia h˜u.u ha. n. 85 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi D- i.nh l´y Cˆosi (Cauchy). Gia’ su.’ : i) f (x) v`a ϕ(x) liˆen tu. c trˆen doa. n [a, b].
ii) f (x) v`a ϕ(x) c´o da. o h`am h˜u.u ha. n trong (a, b).
iii) [f 0(x)]2 + [ϕ0(x)]2 6= 0, ngh˜ıa l`a c´ac da. o h`am khˆong dˆo` ng th`o.i b˘a`ng 0. iv) ϕ(a) 6= ϕ(b). .c diˆe’m ξ ∈ (a, b) sao cho: Khi d´o t`ım du.o. · = (8.14) f (b) − f (a)
ϕ(b) − ϕ(a) f 0(ξ)
ϕ0(ξ) Di.nh l´y Lagrange l`a tru.`o.ng ho. .p riˆeng cu’ a di.nh l´y Cauchy v`ı khi
.c (8.13). Di.nh l´y Rˆon c˜ung l`a tru.`o.ng .p riˆeng cu’ a di.nh l´y Lagrange v´o.i diˆe`u kiˆe.n f (a) = f (b). ϕ(x) = x th`ı t`u. (8.14) thu du.o.
ho. C ´AC V´I DU. V´ı du. 1. Gia’ su.’ P (x) = (x + 3)(x + 2)(x − 1). Ch´u.ng minh r˘a`ng trong khoa’ ng (−3, 1) tˆo` n ta. i nghiˆe.m cu’ a phu.o.ng tr`ınh P 00(ξ) = 0. Gia’ i. Da th´u.c P (x) c´o nghiˆe.m ta. i c´ac diˆe’m x1 = −3, x2 = −2,
x3 = 1. Trong c´ac khoa’ ng (−3, −2) v`a (−2, 1) h`am P (x) kha’ vi v`a
tho’a m˜an c´ac diˆe`u kiˆe.n cu’ a di.nh l´y Rˆon v`a: P (−3) = P (−2) = 0, P (−2) = P (1) = 0. .c diˆe’m ξ1 ∈ (−3, −2); ξ2 ∈ (−2, 1) Do d´o theo di.nh l´y Rˆon, t`ım du.o. sao cho: P 0(ξ1) = P 0(ξ2) = 0. Bˆay gi`o. la. i ´ap du. ng di.nh l´y Rˆon cho doa. n [ξ1, ξ2] v`a h`am P 0(x), ta
.c diˆe’m ξ ∈ (ξ1, ξ2) ⊂ (−3, 1) sao cho P 00(ξ) = 0. la. i t`ım du.o. 86 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n V´ı du. 2. H˜ay x´et xem h`am f (x) = arcsinx trˆen doa. n [−1, +1] c´o
tho’a m˜an di.nh l´y Lagrange khˆong ? Nˆe´u tho’a m˜an th`ı h˜ay t`ım diˆe’m
ξ (xem (8.12)). Gia’ i. H`am f (x) x´ac di.nh v`a liˆen tu. c trˆen [−1, +1]. Ta t`ım f 0(x). √ f 0(x) = → f 0(x) < ∞, x ∈ (−1, 1) 1
1 − x2 (Lu.u ´y r˘a`ng khi x = ±1 da. o h`am khˆong tˆo` n ta. i nhu.ng diˆe’u d´o
. tho’a m˜an diˆe`u kiˆe.n cu’ a di.nh l´y Lagrange !). khˆong a’ nh hu.o.’ ng dˆe´n su.
Nhu. vˆa. y h`am f tho’a m˜an di.nh l´y Lagrange. Ta t`ım diˆe’m ξ. Ta c´o: = arcsin1 − arcsin(−1)
1 − (−1) 1
p1 − ξ2 (cid:17) − (cid:16) − r π
2 π
2 ⇒ = 1 − ⇒ p1 − ξ2 = ⇒ ξ1,2 = ± 2 4
π2 2
π 1
p1 − ξ2 .p n`ay cˆong th´u.c (8.12) tho’a m˜an dˆo´i v´o.i Nhu. vˆa. y trong tru.`o.ng ho. hai diˆe’m. V´ı du. 3. H˜ay kha’ o s´at xem c´ac h`am f (x) = x2 − 2x + 3 v`a ϕ(x) =
x3 − 7x2 + 20x − 5 c´o tho’a m˜an diˆe`u kiˆe.n di.nh l´y Cauchy trˆen doa. n
[1, 4] khˆong ? Nˆe´u ch´ung tho’a m˜an di.nh l´y Cauchy th`ı h˜ay t`ım diˆe’m
ξ. Gia’ i. i) Hiˆe’n nhiˆen ca’ f (x) v`a ϕ(x) liˆen tu. c khi x ∈ [1, 4].
ii) f (x) v`a ϕ(x) c´o da. o h`am h˜u.u ha. n trong (1, 4).
iii) Diˆe`u kiˆe.n th´u. iii) c˜ung tho’a m˜an v`ı: g0(x) = 3x2 − 14x + 20 > 0, x ∈ R. iv) Hiˆe’n nhiˆen ϕ(1) 6= ϕ(4). 87 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi Do d´o f (x) v`a ϕ(x) tho’a m˜an di.nh l´y Cauchy v`a ta c´o = hay = , ξ ∈ (1, 4). f (4) − f (1)
ϕ(4) − ϕ(1) f 0(ξ)
ϕ0(ξ) 11 − 2
27 − 9 2ξ − 2
3ξ2 − 14ξ + 20 .c ξ1 = 2, ξ2 = 4 v`a o.’ dˆay chı’ c´o ξ1 = 2 l`a diˆe’m trong .c cho c´ac h`am f (x) = cos x, T`u. d´o thu du.o.
cu’ a (1, 4). Do d´o: ξ = 2.
V´ı du. 4. Di.nh l´y Cauchy c´o ´ap du. ng du.o.
ϕ(x) = x3 trˆen doa. n [−π/2, π/2] hay khˆong ? Gia’ i. Hiˆe’n nhiˆen f (x) v`a ϕ(x) tho’a m˜an c´ac diˆe`u kiˆe.n i), ii) v`a
iv) cu’ a di.nh l´y Cauchy. Tiˆe´p theo ta c´o: f 0(x) = − sin x; ϕ0(x) = 3x2
v`a ta. i x = 0 ta c´o: f 0(0) = − sin 0 = 0; ϕ0(0) = 0 v`a nhu. vˆa. y
.c tho’a m˜an. Ta
[ϕ0(0)]2 + [f 0(0)]2 = 0. Do d´o diˆe`u kiˆe.n iii) khˆong du.o.
x´et vˆe´ tr´ai cu’ a (8.14): = = 0. f (b) − f (a)
ϕ(b) − ϕ(a) cos(π/2) − cos(−π/2)
(π/2)3 − (−π/2)3 Bˆay gi`o. ta x´et vˆe´ pha’ i cu’ a (8.14). Ta c´o: · = − f 0(ξ)
ϕ0(ξ) sin ξ
3ξ2 Nhu.ng dˆo´i v´o.i vˆe´ pha’ i n`ay ta c´o: (cid:16) − (cid:16) − (cid:17) = ∞. lim
ξ→0 (cid:17) = lim
ξ→0 · lim
ξ→0 sin ξ
3ξ2 sin ξ
ξ 1
3ξ Diˆe`u d´o ch´u.ng to’ r˘a`ng c´ac h`am d˜a cho khˆong tho’a m˜an di.nh l´y Cauchy. B `AI T ˆA. P x2 trˆen doa. n [−1, 1] c´o tho’a m˜an diˆe`u kiˆe.n cu’ a di.nh (Tra’ l`o.i: Khˆong) √
1. H`am y = 1 − 3
l´y Rˆon khˆong ? Ta. i sao ? 88 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n (DS. ξ = π/4) 2. H`am y = 3x2 − 5 c´o tho’a m˜an di.nh l´y Lagrange trˆen doa. n [−2, 0]
(Tra’ l`o.i:
khˆong ? Nˆe´u n´o tho’a m˜an, h˜ay t`ım gi´a tri. trung gian ξ.
C´o)
3. Ch´u.ng minh r˘a`ng h`am f (x) = x + 1/x tho’a m˜an di.nh l´y Lagrange
trˆen doa. n [1/2, 2]. T`ım ξ.
(DS. ξ = 1)
4. Ch´u.ng minh r˘a`ng c´ac h`am f (x) = cos x, ϕ(x) = sin x tho’a m˜an
di.nh l´y Cauchy trˆen doa. n [0, π/2]. T`ım ξ ?
5. Ch´u.ng minh r˘a`ng h`am f (x) = ex v`a ϕ(x) = x2/(1 + x2) khˆong
tho’a m˜an di.nh l´y Cauchy trˆen doa. n [−3, 3].
6. Trˆen du.`o.ng cong y = x3 h˜ay t`ım diˆe’m m`a ta. i d´o tiˆe´p tuyˆe´n v´o.i
du.`o.ng cong song song v´o.i dˆay cung nˆo´i diˆe’m A(−1, −1) v´o.i B(2, 8).
(DS. M(1, 1)) 8.3.2 Khu.’ c´ac da. ng vˆo di.nh. Quy t˘a´c Lˆopitan (L’Hospitale) Chı’ dˆa˜n. Du. .a v`ao ´y ngh˜ıa h`ınh ho. c cu’ a cˆong th´u.c sˆo´ gia h˜u.u ha. n. Trong chu.o.ng II ta d˜a dˆe` cˆa. p dˆe´n viˆe.c khu.’ c´ac da. ng vˆo di.nh. Bˆay gi`o.
ta tr`ınh b`ay quy t˘a´c Lˆopitan - cˆong cu. co. ba’ n dˆe’ khu.’ c´ac da. ng vˆo
di.nh f (x) = 0; Da. ng vˆo di.nh 0/0
Gia’ su.’ hai h`am f (x) v`a ϕ(x) tho’a m˜an c´ac diˆe`u kiˆe.n
ϕ(x) = 0.
i) lim
x→a lim
x→a ii) f (x) v`a ϕ(x) kha’ vi trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m x = a v`a ϕ0(x) 6= 0 trong lˆan cˆa. n d´o, c´o thˆe’ tr`u. ra ch´ınh diˆe’m x = a. iii) Tˆo` n ta. i gi´o.i ha. n (h˜u.u ha. n ho˘a. c vˆo c`ung) = k. lim
x→a f 0(x)
ϕ0(x) 89 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi Khi d´o · lim
x→a = lim
x→a f (x)
ϕ(x) f 0(x)
ϕ0(x) Da. ng vˆo di.nh ∞/∞
Gia’ su.’ f (x) v`a ϕ(x) tho’a m˜an c´ac diˆe`u kiˆe.n ii) v`a iii) cu’ a di.nh l´y trˆen dˆay c`on diˆe`u kiˆe.n i) du.o. i)∗ f (x) = ∞, .c thay bo.’ i diˆe`u kiˆe.n:
ϕ(x) = ∞. lim
x→a lim
x→a
Khi d´o: lim
x→a = lim
x→a f (x)
ϕ(x) f 0(x)
ϕ0(x) Ch´u ´y. Nˆe´u thu.o.ng f 0(x)/ϕ0(x) la. i c´o da. ng vˆo di.nh 0/0 (ho˘a. c
∞/∞) ta. i diˆe’m x = a v`a f 0, ϕ0 tho’a m˜an c´ac diˆe`u kiˆe.n i), ii) v`a iii)
(tu.o.ng ´u.ng i)∗, ii) v`a iii)) th`ı ta c´o thˆe’ chuyˆe’n sang da. o h`am cˆa´p hai,... x→a ϕ(x) = ∞(cid:1) ta C´ac da. ng vˆo di.nh kh´ac
a) Dˆe’ khu.’ da. ng vˆo di.nh 0 · ∞ (cid:0) lim f (x) = 0, lim
x→a biˆe´n dˆo’i t´ıch f (x) · ϕ(x) th`anh: i) (da. ng 0/0) ii) (da. ng ∞/∞). f (x)
1/ϕ(x)
ϕ(x)
1/f(x) ϕ(x) = ∞) b) Dˆe’ khu.’ da. ng vˆo di.nh ∞ − ∞
Ta biˆe´n dˆo’i f (x) − ϕ(x) (trong d´o lim
x→a f (x) = ∞, lim
x→a th`anh t´ıch i − f (x) − ϕ(x) = f (x)ϕ(x)h 1
ϕ(x) 1
f (x) ho˘a. c th`anh t´ıch da. ng i f (x) − ϕ(x) = f (x)h1 − ϕ(x)
f (x) 90 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n ho˘a. c f (x) − ϕ(x) = ϕ(x)h − 1i. f (x)
ϕ(x) c) Da. ng vˆo di.nh 00, ∞0, 1∞
Khi t´ınh gi´o.i ha. n cu’ a h`am da. ng F (x) = [f (x)]ϕ(x) thˆong thu.`o.ng
ta g˘a. p c´ac da. ng vˆo di.nh 00, ∞0 ho˘a. c 1∞. Trong nh˜u.ng tru.`o.ng ho.
.p
n`ay ta c´o thˆe’ biˆe´n dˆo’i F (x) dˆe’ du.a vˆe` da. ng vˆo di.nh 0 · ∞ d˜a n´oi trong
1) nh`o. ph´ep biˆe´n dˆo’i F (x) = [f (x)]ϕ(x) = eln[f (x)]ϕ(x) = eϕ(x)lnf (x) v`a do t´ınh liˆen tu. c cu’ a h`am m˜u ta s´e c´o: [f (x)]ϕ(x) = elim[ϕ(x)·lnf (x)] lim
x→a Ch´u ´y. Ta lu.u ´y r˘a`ng m˘a. c d`u quy t˘a´c Lˆopitan l`a mˆo. t cˆong cu.
ma. nh de’ t´ınh gi´o.i ha. n nhu.ng n´o khˆong thˆe’ thay to`an bˆo. c´ac phu.o.ng
ph´ap t´ınh gi´o.i ha. n d˜a x´et trong chu.o.ng II. Diˆe`u d´o du.o.
.c ch´u.ng to’
trong v´ı du. 7 sau dˆay. C ´AC V´I DU. V´ı du. 1. T´ınh lim
x→1 x2 − 1 + lnx
ex − e Gia’ i. Ta c´o vˆo di.nh da. ng “0/0”. ´Ap du. ng quy t˘a´c L’Hospital ta .c
thu du.o. 2x + . N lim
x→1 = lim
x→1 = lim
x→1 x2 − 1 + lnx
ex − e (x2 − 1 + lnx)0
(ex − e)0 1
x
ex = 3
e V´ı du. 2. T´ınh lim
x→+∞ xn
ex 91 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi Gia’ i. Ta c´o vˆo di.nh da. ng “∞/∞”. ´Ap du. ng quy t˘a´c L’Hospital n .c
lˆa` n ta thu du.o. lim
x→∞ = lim
x→1 = · · · = lim
x→1 = lim
x→1 xn
ex nxn−1
ex n(n − 1)xn−2
ex n(n − 1) · · · 2 · 1
ex = 0. N = lim
x→1 n!
ex V´ı du. 3. T´ınh lim
x→0+0 xlnx.
Gia’ i. Ta c´o vˆo di.nh da. ng “0 · ∞”. Nhu.ng xlnx = lnx
1
x v`a ta thu du.o. .c vˆo di.nh da. ng “∞/∞”. Do d´o 0 = lim
x→0+0
(cid:17) x = 0. N lim
x→0+0 xlnx = lim
x→0+0 = − lim
x→0+0 (cid:16) − (lnx)0
1
x 1
x
1
x2 xx. V´ı du. 4. T´ınh lim
x→0+0 Gia’ i. O .’ dˆay ta c´o vˆo di.nh da. ng “00”. Nhu.ng xx = exlnx .c vˆo di.nh da. ng 0 · ∞ o.’ sˆo´ m˜u. Trong v´ı du. 3 ta d˜a thu v`a ta thu du.o.
.c
du.o. (xlnx) = 0, lim
x→0+0 xlnx lim
x→0+0 do d´o 1
ex −1−x exlnx = e = e0 = 1. N lim
x→0+0 xx = lim
x→0+0 (cid:0)1 + x2(cid:1) V´ı du. 5. T´ınh lim
x→0 92 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 1 ln(1+x2 )
ex −1−x ex −1−x = e Gia’ i. O .’ dˆay ta c´o vˆo di.nh da. ng 1∞. Nhu.ng (cid:0)1 + x2(cid:1) .c vˆo di.nh da. ng “0/0”. ´Ap du. ng = lim
x→0 lim
x→0 v`a o.’ sˆo´ m˜u cu’ a l˜uy th`u.a ta thu du.o.
.c
quy t˘a´c L’Hospital ta thu du.o.
2x
1 + x2
ex − 1 ln(1 + x2)
ex − 1 − x 2x
(ex − 1)(1 + x2) 2 cos x. = = 2. N = lim
x→0 = lim
x→0
2
ex(1 + x2) + (ex − 1)2x 2
1 (cid:0)tgx(cid:1) V´ı du. 6. T´ınh lim
x→ π
2 2ln tgx
1/ cos x 2 cos x = e2 cos xln tgx = e Gia’ i. Ta c´o vˆo di.nh da. ng “∞0”. Nhu.ng (cid:0)tgx(cid:1) .c vˆo di.nh da. ng “∞/∞”. ´Ap du. ng v`a o.’ sˆo´ m˜u cu’ a l˜uy th`u.a ta thu du.o.
quy t˘a´c L’Hospital ta c´o 1
cos x
tg2x lim
x→ π
2 = 2 lim
x→ π
2 = 2 lim
x→ π
2 2ln tgx
1
cos x − cos x = 0. = 2 lim
x→ π
2 = lim
x→ π
2 2tgx · 1
cos2 x · tgx
+ sin x
cos2 x
sin x
cos2 x
1
cos2 x 2 cos x·ln tgx lim
x→ π
2 2 cos x = e Do d´o = e0 = 1. N (cid:0)tgx(cid:1) lim
x→ π
2 V´ı du. 7. Ch´u.ng minh r˘a`ng gi´o.i ha. n = 0 1) lim
x→0 x2 sin(1/x)
sin x 93 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi = 1 2) .c theo quy t˘a´c L’Hospital. H˜ay t´ınh c´ac gi´o.i ha. n x − sin x
lim
x + sin x
x→∞
khˆong thˆe’ t`ım du.o.
d´o. .c v`ı ty’ sˆo´ c´ac da. o Gia’ i. 1) Quy t˘a´c L’Hospital khˆong ´ap du. ng du.o. h`am [2x sin(1/x) − cos(1/x)]/ cos x khˆong c´o gi´o.i ha. n khi x → 0. .c tiˆe´p gi´o.i ha. n n`ay. Ta t´ınh tru. = 1 · 0 = 0. x sin lim
x→0 = lim
x→0 · lim
x→0 x2 sin(1/x)
sin x x
sin x 1
x
.c v`ı ty’ sˆo´ c´ac da.o h`am 2) Quy t˘a´c L’Hospital khˆong ´ap du. ng du.o. = tg2(x/2) 1 − cos x
1 + cos x khˆong c´o gi´o.i ha. n khi x → ∞. .c tiˆe´p gi´o.i ha. n n`ay Ta t´ınh tru. = 1 v`ı | sin x| 6 1. lim
x→∞ = lim
x→∞ x − sin x
x + sin x [1 − (sin x)/x]
[1 + (sin x)/x] lim
x→a Nhu. o.’ phˆa` n dˆa` u cu’ a tiˆe´t n`ay d˜a n´oi, quy t˘a´c L’Hospital l`a mˆo. t
cˆong cu. ma. nh dˆe’ t`ım gi´o.i ha. n nhu.ng diˆe`u d´o khˆong c´o ngh˜ıa l`a n´o c´o
thˆe’ thay cho to`an bˆo. c´ac phu.o.ng ph´ap t`ım gi´o.i ha. n. Cˆa` n lu.u ´y r˘a`ng
f (x)
quy t˘a´c L’Hospital chı’ l`a diˆe`u kiˆe.n du’ dˆe’ tˆo` n ta. i gi´o.i ha. n:
g(x)
ch´u. khˆong pha’ i l`a diˆe`u kiˆe.n cˆa` n. B `AI T ˆA. P ´Ap du. ng quy t˘a´c L’Hospital dˆe’ t´ınh gi´o.i ha. n: . ) (DS. 1. lim
x→2 16
13 (DS. . am−n) 2. lim
x→a x4 − 16
x3 + 5x2 − 6x − 16
xm − am
xn − an m
n 94 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n . 3. (DS. 2) lim
x→0 . (DS. ) 4. lim
x→0 a2
b2 . (DS. 2) 5. lim
x→0 . (DS. 0) 6. lim
x→0 . ) (Ds. − 7. lim
x→∞ 1
2 . (DS. 0) 8. lim
x→∞ e2x − 1
sin x
1 − cos ax
1 + cos bx
ex − e−x − 2x
x − sin x
ln(1 + x2)
cos 3x − e−x
e1/x2 − 1
2arctgx2 − π
2x + 1
3x2 + x − 1 . (DS. −2) 9. lim
x→∞ ln(1 + x2)
ln[(π/2) − arctgx] √ 10. . (DS. −1) lim
x→∞ . (DS. +∞) 11. lim
x→∞ x2 − 1
x
x
ln(1 + x) . 12. (DS. 1) lim
x→+0 ln sin x
ln sin 5x cotg(x − a). arcsin (DS. 1/a) x − a
a
(π − 2arctgx)lnx. (DS. 0) 14. 13. lim
x→a
lim
x→∞ (a1/x − 1)x, a > 0. (DS. lna) 15. lim
x→∞ (2 − x)tg πx
2 . (DS. e2/π) 16. lim
x→1 h − 17. i. (DS. −1) lim
x→1 18. x
1
lnx
lnx
(x − x2ln(1 + 1/x)). (Ds. 1/2) lim
x→∞ (cid:16) 19. − cotg2x(cid:17). (DS. 2/3) lim
x→0 1
x2 x1/ln(ex−1). (DS. e) 20. lim
x→0 tgx 95 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi 1/ sin x 21. . (DS. 1) (cid:0)cotgx(cid:1) lim
x→0+0 (cid:16) (cid:17) 22. . (DS. e−1/30) lim
x→0 2 + 23. 5
√
9 + x
cotg2x. (DS. e−1/2) (cid:0) cos x(cid:1) lim
x→0 1/lnx. 24. (DS. 1) (cid:0)ln2x(cid:1) lim
x→0+0 1/tg2x. 25. (DS. e) (cid:0)1 + sin2 x(cid:1) lim
x→0 1/lnx. tgx. 26. (DS. e−1) (cid:0)cotgx(cid:1) lim
x→0+0 (DS. 1) 27. (cid:0) sin x(cid:1) lim
x→π/2 28. . (DS. −2) lim
x→0 e+x − e−x − 2x
sin x − x e−x − 1 + x − x2
2 29. . ) (DS. − lim
x→0 1
6 . ) (DS. − 30. lim
x→0 1
2 31. . ) (DS. lim
x→0 ex3 − 1
e−x − 1 + x4
sin 2x
2x − 1 − xln2
(1 − x)m − 1 + mx ln22
m(m − 1) 1/x. (cid:16) (DS. e− 2
π ) 32. arccosx(cid:1) lim
x→0 33. , α > 0. (DS. 0) lim
x→∞ 34. , 0 < a 6= 1. (DS. 0) lim
x→∞ 2
π
lnx
xα
xm
ax 35. . ) (DS. lim
x→0+0 h ) − cotg2xi. (DS. 36. lim
x→0 1
2
2
3 tg2x. cotgx. 37. (DS. e−1) ln sin x
ln(1 − cos x)
1
x2
(cid:0)tgx(cid:1) lim
x→ π
4 −0 (DS. 1) 38. (cid:0)tgx(cid:1) lim
x→ π
2 8.3.3 Cˆong th´u.c Taylor
Gia’ su.’ h`am f (x) x´ac di.nh trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m x0 v`a n lˆa` n
kha’ vi ta. i diˆe’m x0 th`ı 96 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n f (x) = f (x0) + (x − x0) + (x − x0)2 + · · · + f 0(x0)
1! f 00(x0)
2! + (x − x0)n + o((x − x0)n) f (n)(x0)
n! khi x → x0 hay: n
X
k=0 f (x) = (8.15) (x − x0)k + o((x − x0)n), x → x0. f (k)(x0)
k! Da th´u.c n
X
k=0 (8.16) Pn(x) = (x − x0)k f (k)(x0)
k! du.o. .c go. i l`a da th´u.c Taylor cu’ a h`am f (x) ta. i diˆe’m x0, c`on h`am: Rn(x) = f (x) − Pn(x) du.o. .c go. i l`a sˆo´ ha. ng du. hay phˆa` n du. th´u. n cu’ a cˆong th´u.c Taylor.
Cˆong th´u.c (8.15) du.o. .c go. i l`a cˆong th´u.c Taylor cˆa´p n dˆo´i v´o.i h`am
f (x) ta. i lˆan cˆa. n cu’ a diˆe’m x0 v´o.i phˆa` n du. da. ng Peano (n´o c˜ung c`on
.c go. i l`a cˆong th´u.c Taylor di.a phu.o.ng). Nˆe´u h`am f (x) c´o da. o h`am
du.o.
dˆe´n cˆa´p n th`ı n´o c´o thˆe’ biˆe’u diˆe˜n duy nhˆa´t du.´o.i da. ng: n
X
k=0 f (x) = ak(x − x0)k + o((x − x0)n), x → x0 .c t´ınh theo cˆong th´u.c: v´o.i c´ac hˆe. sˆo´ ak du.o. , k = 0, 1, . . . , n. ak = f (k)(x0)
k! 97 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi Nˆe´u x0 = 0 th`ı (8.15) c´o da. ng n
X
k=0 f (x) = xk + o(xn), x → 0 (8.17) f (k)(0)
k! v`a go. i l`a cˆong th´u.c Macloranh (Maclaurin). Sau dˆay l`a cˆong th´u.c Taylor ta. i lˆan cˆa. n diˆe’m x0 = 0 cu’ a mˆo. t sˆo´ h`am so. cˆa´p n
P
k=0 I. ex = + o(xn) xk
k! + · · · + + + o(x2n+2) II. sin x = x − (−1)nx2n+1
(2n + 1)! n
X
k=0 = + o(x2n+2) x3
x5
3!
5!
(−1)k x2k+1
(2k + 1)! n
P
k=0 + o(x2n+1) III. cos x = (−1)k x2k
(2k!) IV. (1 + x)α = 1 + xk + o(xn) α(α − 1) . . . (α − k + 1)
k! n
X
k=1
n
X
k=1 ! xk + o(xn) = 1 + α
k α nˆe´u α ∈ R, = k α(α − 1) . . . (α − k + 1)
k! nˆe´u α ∈ N. C k
α
.p riˆeng: n
P
k=0 = (−1)kxk + o(xn), IV1. Tru.`o.ng ho.
1
1 + x n
P
k=0 = xk + o(xn). IV2. 1
1 − x 98 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n n
X
k=1 V. ln(1 + x) = xk + o(xn). (−1)k−1
k n
X
k=1 Phu.o.ng ph´ap khai triˆe’n theo cˆong th´u.c Taylor
Nhu. vˆa. y, dˆe’ khai triˆe’n h`am f (x) theo cˆong th´u.c Taylor ta pha’ i ´ap + o(xn). ln(1 − x) = − xk
k du. ng cˆong th´u.c f (x) = Tn(x) + Rn+1(x), n
X
k=0 Tn(x) = ak(x − x0)k, · (8.18) ak = 1) Phu.o.ng ph´ap tru. .c tiˆe´p: du. f (k)(x0)
k!
.a v`ao cˆong th´u.c (8.18). Viˆe.c su.’
du. ng cˆong th´u.c (8.18) dˆa˜n dˆe´n nh˜u.ng t´ınh to´an rˆa´t cˆo` ng kˆe`nh m˘a. c
d`u n´o cho ta kha’ n˘ang nguyˆen t˘a´c dˆe’ khai triˆe’n. 2) Phu.o.ng ph´ap gi´an tiˆe´p: du. .a v`ao c´ac khai triˆe’n c´o s˘a˜n I-V sau
.c hiˆe.n khi d˜a biˆe´n dˆo’i so. bˆo. h`am d˜a cho v`a lu.u ´y dˆe´n c´ac quy t˘a´c thu.
c´ac ph´ep to´an trˆen c´ac khai triˆe’n Taylor. Nˆe´u f (x) = ak(x − x0)k + o((x − x0)n) n
X
k=0
n
X
k=0 g(x) = bk(x − x0)k + o((x − x0)n) th`ı n
P
k=0 a) f (x) + g(x) = (ak + bk)(x − x0)k + o((x − x0)n); 99 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi n
P
k=0 b) f (x)g(x) = ck(x − x0)k + o((x − x0)n) k
X
p=0 ck = apbk−p j
(bk(x − x0)k − x0i n
P
j=0 n
P
k=0 n(cid:17) c) F (x) = f [g(x)] = ajh n
P
k=0 bk(x − x0)k − x0(cid:1) +o(cid:16)(cid:0) 3) Dˆe’ khai triˆe’n c´ac phˆan th´u.c h˜u.u ty’ theo cˆong th´u.c Taylor thˆong
thu.`o.ng ta biˆe’u diˆe˜n phˆan th´u.c d´o du.´o.i da. ng tˆo’ng cu’ a da th´u.c v`a c´ac
phˆan th´u.c co. ba’ n (tˆo´i gia’ n !) rˆo` i ´ap du. ng VI1, IV2. .ng gi´ac thˆong thu.`o.ng biˆe´n dˆo’i 4) Dˆe’ khai triˆe’n t´ıch c´ac h`am lu.o. t´ıch th`anh tˆo’ng c´ac h`am. 5) Nˆe´u cho tru.´o.c khai triˆe’n da. o h`am f 0(x) theo cˆong th´u.c Taylor
.c hiˆe.n nhu. sau. th`ı viˆe.c t`ım khai triˆe’n Taylor cu’ a h`am f (x) du.o. .c thu. Gia’ su.’ cho biˆe´t khai triˆe’n f 0(x) = bk(x − x0)k + o((x − x0)n), n
X
k=0
f (k+1)(x0)
k! · bk = Khi d´o tˆo` n ta. i f (n+1)(x0) v`a do d´o h`am f (x) c´o thˆe’ biˆe’u diˆe˜n du.´o.i
da. ng n+1
X
k=0 f (x) = ak(x − x0)k + o((x − x0)n+1) n
X
k=0 = f (x0) + ak+1(x − x0)k+1 + o((x − x0)n+1) trong d´o · · = = ak+1 = f (k+1)(x0)
(k + 1)! f (k+1)(x0)
k! 1
k + 1 bk
k + 1 100 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n Do d´o n
X
k=0 (8.19) f (x) = f (x0) + (x − x0)k+1 + o((x − x0)n+1) bk
k + 1 trong d´o bk l`a hˆe. sˆo´ cu’ a da th´u.c Taylor dˆo´i v´o.i h`am f 0(x). C ´AC V´I DU.
V´ı du. 1. Khai triˆe’n h`am f (x) theo cˆong th´u.c Maclaurin dˆe´n sˆo´ ha. ng
o(xn), nˆe´u 2) f (x) = ln 1) f (x) = (x + 5)e2x; 3 + x
2 − x .c
Gia’ i 1) Ta c´o f (x) = xe2x + 5e2x. ´Ap du. ng I ta thu du.o. n
X
k=0 f (x) = x(cid:16) + o(xn−1)(cid:17) + 5(cid:16) + o(xn)(cid:17) 2kxk
k! 2kxk
k! n−1
X
k=0
2k
k! n−1
X
k=0 n
X
k=0 xk+1 + xk + o(xn). = 5 · 2k
k! n−1
P
k=0 n
P
k=1 = xk nˆen ta c´o V`ı 2kxk+1
k! 2k−1
(k − 1)! h f (x) = 5 + + ixk + o(xn) 2k−1
(k − 1)! 5 · 2k
k! n
X
k=1
2k−1
k! n
X
k=0 (k + 10)xk + o(xn). = 2) T`u. d˘a’ ng th´u.c (cid:17) f (x) = ln + ln(cid:16)1 + (cid:17) − ln(cid:16)1 − 3
2 x
3 x
2 .c
v`a V ta thu du.o. n
X
k=1 (cid:16) f (x) = ln + + (cid:17)xk + o(xn). N 3
2 1
k 1
2k (−1)k−1
3k 101 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi V´ı du. 2. Khai triˆe’n h`am f (x) theo cˆong th´u.c Taylor ta. i lˆan cˆa. n diˆe’m
x0 = −1 dˆe´n sˆo´ ha. ng o((x + 1)2n) nˆe´u √ · f (x) = 3x + 3
3 − 2x − x2 Gia’ i. Ta c´o − 1
2 . (cid:17) f (x) = (x + 1)(cid:16)1 − = 3
2 (x + 1)2
4 3(x + 1)
p4 − (x + 1)2 n−1
X
k=1 − (−1)k (x + 1)2k (x + 1) + (x + 1) + o((x + 1)2n) f (x) = 3
2 3
2 4k .c
´Ap du. ng cˆong th´u.c IV ta thu du.o.
1
2
k trong d´o (cid:16) − (cid:17)(cid:16) − − 1(cid:17) . . . (cid:16) − − (k − 1)(cid:17) − 1
2 1
2 1
2 (−1)k = (−1)k k! 1
2
k · = (2k − 1)!!
2kk! Do d´o n−1
X
k=1 (x + 1) + (x + 1)2k+1 + o((x + 1)2n). N f (x) = 3
2 3(2k − 1)!!
23k+1k! V´ı du. 3. Khai triˆe’n h`am f (x) theo cˆong th´u.c Taylor ta. i lˆan cˆa. n diˆe’m
x0 = 2 dˆe´n sˆo´ ha. ng o((x − 2)n), nˆe´u f (x) = ln(2x − x2 + 3). Gia’ i. Ta biˆe’u diˆe˜n 2x − x2 + 3 = (3 − x)(x + 1) = [1 − (x − 2)][3 + (x − 2)] i. = 3[1 − (x − 2)]h1 + x − 2
3 102 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n T`u. d´o suy ra r˘a`ng i f (x) = ln3 + ln[1 − (x − 2)] + lnh1 + x − 2
3 n
X
k=1 f (x) = ln3 − .c
v`a ´ap du. ng cˆong th´u.c V ta thu du.o.
(−1)k−1 (x − 2)k (x − 2)k + + o((x − 2)n) k3k 1
k n
X
k=1
n
X
k=1 h − 1i + o((x − 2)n). N = ln3 + (−1)k−1
3k (x − 2)k
k V´ı du. 4. Khai triˆe’n h`am f (x) = ln cos x theo cˆong th´u.c Maclaurin
dˆe´n sˆo´ ha. ng ch´u.a x4. + + o(x4)i = ln(1 + t), ln(cos x) = lnh1 − .c
Gia’ i. ´Ap du. ng III ta thu du.o.
x2
2 x4
24 trong d´o ta d˘a. t + + o(x4). t = − x2
2 x4
24
Tiˆe´p theo ta ´ap du. ng khai triˆe’n V 2(cid:17) + o(t2) ln(cos x) = ln(1 + t) = t − t2
2 2(cid:17) + + = (cid:16) − + o(x4) − + o(x4)(cid:1) (cid:0) − x2
2 x4
24 1
2 x2
2 x4
4 + + o(x4)(cid:1) − − + = − + o(x4) = − + o(x4). N + o(cid:16)(cid:0) −
x2
2 x2
2
x4
24 x4
24
x4
8 x2
2 x4
12 V´ı du. 5. Khai triˆe’n h`am f (x) = ex cos x theo cˆong th´u.c Maclaurin
dˆe´n sˆo´ ha. ng ch´u.a x3. Gia’ i. Khai triˆe’n cˆa` n t`ım pha’ i c´o da. ng 3
X
k=0 ex cos x = akxk + o(x3). 103 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi V`ı x cos x = x + 0(x), (x cos x)k = xk + o(xk), k = 1, 2, . . . nˆen trong cˆong th´u.c n
X
k=0 ew = + o(wn), w = x cos x wk
k! ta cˆa` n lˆa´y n = 3. Ta c´o w = x cos x = x − + o(x4) x3
2! w2 = x2 + o(x3), w3 = x3 + o(x3) v`a do d´o 3
X
k=0 + o(w3) ex cos x = wk
k! + o(x4) + = 1 + x − (cid:0)x2 + 0(x3)(cid:1) + (cid:0)x3 + o(x3)(cid:1) + 0(x3) 1
2 1
3! x3 + o(x3). N = 1 + x + x3
2!
1
x2 −
2 1
3 V´ı du. 6. Khai triˆe’n theo cˆong th´u.c Maclaurin dˆe´n o(x2n+1) dˆo´i v´o.i
c´ac h`am 2) arc sin x. 1) arctgx,
Gia’ i. 1) V`ı n
X
k=0 (arctgx)0 = (−1)kx2k + o(x2n+1) = 1
1 + x2 nˆen theo cˆong th´u.c (8.19) ta c´o n
X
k=0 + o(x2n+2). arctgx = (−1)k x2k+1
(2k + 1) .c
V´o.i n = 2 ta thu du.o. arctgx = x − + + o(x6) x3
3 x5
5 104 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n 2) Ta c´o n
X
k=1 − √ (arcsinx)0 = = 1 + (−1)k x2k + o(x2n+1) 1
1 − x2 1
2
k n
X
k=1 (−1)k (2k − 1)!! x2k + o(x2n+1). = 1 + 2kk! T`u. d´o ´ap du. ng cˆong th´u.c (8.19) ta c´o n
X
k=0 arc sin x = x + x2k+1 + o(x2n+2). (2k − 1)!!
2kk!(2k + 1) .c
V´o.i n = 2 ta thu du.o. arc sin x = x + x3 + x5 + o(x6). N 1
6 3
40 V´ı du. 7. Khai triˆe’n h`am f (x) = tgx theo cˆong th´u.c Maclaurin dˆe´n
o(x5). .c
Gia’ i. Ta s˜e d`ung phu.o.ng ph´ap hˆe. sˆo´ bˆa´t di.nh m`a nˆo. i dung du.o. thˆe’ hiˆe.n trong l`o.i gia’ i sau dˆay. V`ı tgx l`a h`am le’ v`a tgx = x + o(x) nˆen tgx = x + a3x3 + asx5 + o(x6). Ta su.’ du. ng cˆong th´u.c sin x = tgx · cos x v`a c´ac khai triˆe’n II v`a III ta c´o + + + 0(x5)(cid:17) x − + o(x6) = (cid:0)x + a3x3 + a5x5 + o(x6)(cid:1)(cid:16)1 − x5
5 x4
4! x3
x2
3
2!
.c
Cˆan b˘a`ng c´ac hˆe. sˆo´ cu’ a x3 v`a x5 o.’ hai vˆe´ ta thu du.o. − = − + a3 1
6 1
2
− = + a5 1
5! 1
4! a3
2! 105 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi . Do d´o T`u. d´o suy ra a3 = , a5 = 1
3 2
15 tgx = x + + + o(x6). N x3
3 2x5
15 V´ı du. 8. ´Ap du. ng cˆong th´u.c Maclaurin dˆe’ t´ınh c´ac gi´o.i ha. n sau: 2 − cos x
x3 sin x e− x2 · 1) , 2) lim
x→0 lim
x→0 sin x − x
x3 + o(x4) − x x − Gia’ i. 1) ´Ap du. ng khai triˆe’n dˆo´i v´o.i h`am sin x v´o.i n = 2 ta c´o
x3
3! lim
x→0 sin x − x
x3 · + 0 = − = − + lim
x→0 = lim
x→0
1
3! x3
o(x4)
x3 = − 1
3! 1
3! 2) ´Ap du. ng c´ac khai triˆe’n ba’ ng dˆo´i v´o.i et, cos t, sin t cho tru.`o.ng
.p n`ay ta c´o ho. 2 − cos x
x3 sin x − + 1 − + o(x4) − 1 + + 0(x4) e− x2 x2
2 x4
8 x4
24 lim
x→0 = lim
x→0 x2
2
x3(x + 0(x)) − − + + 0(x4) x4
8 x4
24 1
8 1
24 o(x4)
x4 = lim
x→0 = lim
x→0 x4 + 0(x4) 1 + 0(x4)
x4 − + 0 1
8 = · N = 1
24
1 + 0 1
12 B `AI T ˆA. P Khai triˆe’n c´ac h`am theo cˆong th´u.c Maclaurin dˆe´n o(xn) (1-8) n
P
k=0 1. f (x) = . (DS. xk + o(xn)) 1
3x + 4 (−1)k 3k
4k+1 106 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n k! n
P
k=0 √ . 2. f (x) = (DS. (−1)k 2k(2k−1)!! xk + o(xn)) 1
1 + 4x n
P
k=0 (DS. (k + 1)xk + o(xn)) . 3. f (x) = 1
(1 − x)2 n
P
k=1 4. f (x) = ln . + (DS. ln xk + o(xn)) 2 − 3x
2 + 3x 2
3 (−4)k − 9k
k6k 5. f (x) = ln(x2 + 3x + 2). n
P
k=1 (1 + 2−k)xk + o(xn)) (DS. ln2 + (−1)k−1
k n
P
k=1 6. f (x) = ln(2 + x − x2). (DS. ln2 + xk + o(xn)) (−1)k−1 − 2−k
k . 7. f (x) = 1 − 2x2
2 + x − x2 n
P
k=1 + xk + o(xn)) (DS. 1
2 (−1)k+1 − 7 · 2−(k+1)
3 n
P
k=1 + . (DS. 8. f (x) = (cid:2)(−1)k2−(k+1) − 1(cid:3)xk + o(xn)). 3x2 + 5x − 5
x2 + x − 2 5
2 Khai trˆe’n h`am theo cˆong th´u.c Maclaurin dˆe´n 0(x2n+1) (9-13) (DS. x2k + o(x2n+1)) 9. f (x) = sin2 x cos2 x. (−1)k+124k−3
(2k)! n
P
k=0 (DS. (32k−1 + 1)x2k + o(x2n+1)) 10. f (x) = cos3 x. n
P
k=1
3(−1)k
4(2k)!
3
4 cos 3x + cos x. Chı’ dˆa˜n. cos3 x = 1
4 11. f (x) = cos4 x + sin4 x. n
P
k=1 x2k + 0(x2k+1)) (DS. 1 + (−1)k 42k
(2k)! + cos 4x. Chı’ dˆa˜n. Ch´u.ng minh r˘a`ng cos4 x + sin4 x = 3
4 1
4 12. f (x) = cos6 x + sin6 x. n
P
k=1 x2k + o(x2n+1)) (DS. 1 + 3(−1)k42k−1
2(2k)! 107 8.3. C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi 13. f (x) = sin x sin 3x. n
P
k=0 (DS. (1 − 22k)x2k + o(x2n+1)). (−1)k22k−1
(2k)! Khai triˆe’n h`am theo cˆong th´u.c Taylor trong lˆan cˆa. n diˆe’m x0 dˆe´n o((x − x0)n) (14-20) n
P
k=0 √ 14. f (x) = (DS. (x − 1)k + o((x − 1)n)) x, x0 = 1. 1
2
k 15. f (x) = (x2 − 1)e2x, x0 = −1. n
P
k=1 (DS. (x + 1)k + o((x + 1)n)) e−22k−2(k − 5)
(k − 1)! 16. f (x) = ln(x2 − 7x + 12), x0 = 1. n
P
k=1 (DS. ln6 − (x − 1)k + o(x − 1)n)) 2−k + 3−k
k 17. f (x) = ln , x0 = 2. (x − 1)x−2
3 − x n
P
k=2 (cid:16) + (DS. (x − 2) + (cid:17)(x − 2)k + o((x − 2)n)) 1
k (−1)k
k − 1 n
P
k=2 (DS. (x − 2)k + o((x − 2)n)) 18. f (x) = , x0 = 2. (x − 2)2
3 − x 19. f (x) = , x = 3. (DS. 3 + (−1)k(x − 3)k + o((x − 3)n)) 20. f (x) = , x0 = 2. (DS. (x + 2)k + o((x + 2)n)). (−1)k(k − 1)
3k x2 − 3x + 3
x − 2
n
P
k=2
x2 + 4x + 4
x2 + 10x + 25
n
P
k=2 ´Ap du. ng cˆong th´u.c Maclaurin dˆe’ t´ınh gi´o.i ha. n . (DS. 2) 21. lim
x→0 ex − e−x − 2x
x − sin x . (DS. 0) 22. lim
x→0 tgx + 2 sin x − 3x
x4 108 Chu.o.ng 8. Ph´ep t´ınh vi phˆan h`am mˆo. t biˆe´n . (DS. 1) 23. lim
x→0 ex − e−x − 2
x2 2 (cid:16) − (cid:17). (DS. 0) 24. lim
x→0 ) . (DS. − 25. lim
x→0 1
12 1
1
x
sin x
cos x − e− x2
x4
√
1 + x2 cos x 1 − ) . (DS. 26. lim
x→0 x4 1
3 (cid:17) ln(cid:16) cos x + . ) (DS. − 27. lim
x→0 x2
2
x(sin x − x) 1
4 √
sin(sin x) − x 3 1 − x2 ) . (DS. 28. lim
x→0 x5 19
90 9.1 D- a. o h`am riˆeng . . . . . . . . . . . . . . . . . 110
9.1.1 D- a. o h`am riˆeng cˆa´p 1 . . . . . . . . . . . . . 110
.p . . . . . . . . . . . . 111
9.1.2 D- a. o h`am cu’ a h`am ho.
9.1.3 H`am kha’ vi
. . . . . . . . . . . . . . . . . . 111
9.1.4 D- a. o h`am theo hu.´o.ng . . . . . . . . . . . . . 112
9.1.5 D- a. o h`am riˆeng cˆa´p cao . . . . . . . . . . . . 113
9.2 Vi phˆan cu’ a h`am nhiˆe`u biˆe´n . . . . . . . . . 125 9.2.2 9.2.1 Vi phˆan cˆa´p 1 . . . . . . . . . . . . . . . . . 126
´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung . . . . . 126
9.2.3 C´ac t´ınh chˆa´t cu’ a vi phˆan . . . . . . . . . . 127 9.2.4 Vi phˆan cˆa´p cao . . . . . . . . . . . . . . . 127
9.2.5 Cˆong th´u.c Taylor . . . . . . . . . . . . . . . 129 9.3 Cu. 9.2.6 Vi phˆan cu’ a h`am ˆa’n . . . . . . . . . . . . . 130
.c tri. cu’ a h`am nhiˆe`u biˆe´n . . . . . . . . . 145 9.3.1 Cu. .c tri.
. . . . . . . . . . . . . . . . . . . . 145
.c tri. c´o diˆe`u kiˆe.n . . . . . . . . . . . . . 146
9.3.2 Cu.
9.3.3 Gi´a tri. l´o.n nhˆa´t v`a b´e nhˆa´t cu’ a h`am . . . . 147 110 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n 9.1.1 D- a. o h`am riˆeng cˆa´p 1
Gia’ su.’ w = f (M), M = (x, y) x´ac di.nh trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m
M(x, y). Ta. i diˆe’m M ta cho biˆe´n x sˆo´ gia t`uy ´y ∆x trong khi vˆa˜n gi˜u.
gi´a tri. cu’ a biˆe´n y khˆong dˆo’i. Khi d´o h`am f (x, y) nhˆa. n sˆo´ gia tu.o.ng
´u.ng l`a ∆xw = f (x + ∆x, y) − f (x, y) go. i l`a sˆo´ gia riˆeng cu’ a h`am f (x, y) theo biˆe´n x ta. i diˆe’m M(x, y). .ng Tu.o.ng tu. . da. i lu.o. ∆yw = f (x, y + ∆y) − f (x, y) go. i l`a sˆo´ gia riˆeng cu’ a h`am f (x, y) theo biˆe´n y ta. i diˆe’m M(x, y).
D- i.nh ngh˜ıa 9.1.1 1. Nˆe´u tˆo` n ta. i gi´o.i ha. n h˜u.u ha. n lim
∆x→0 = lim
∆x→0 f (x + ∆x, y) − f (x, y)
∆xw
∆x
∆x
.c go. i l`a da. o h`am riˆeng cu’ a h`am f (x, y) theo biˆe´n .c chı’ bo.’ i mˆo. t trong c´ac k´y hiˆe.u th`ı gi´o.i ha. n d´o du.o.
x ta. i diˆe’m (x, y) v`a du.o. x(x, y), w0
f 0
x. , , ∂w
∂x ∂f (x, y)
∂x 111 9.1. D- a. o h`am riˆeng 2. Tu.o.ng tu. .: nˆe´u tˆo` n ta. i gi´o.i ha. n lim
∆y→0 = lim
∆y→0 f (x, y + ∆y) − f (x, y)
∆yw
∆y
∆y
.c go. i l`a da. o h`am riˆeng cu’ a h`am f (x, y) theo biˆe´n .c chı’ bo.’ i mˆo. t trong c´ac k´y hiˆe.u th`ı gi´o.i ha. n d´o du.o.
y ta. i diˆe’m M(x, y) v`a du.o. y(x, y), w0
f 0
y. , , ∂w
∂y ∂f (x, y)
∂y T`u. di.nh ngh˜ıa suy r˘a`ng da. o h`am riˆeng cu’ a h`am hai biˆe´n theo biˆe´n
x l`a da. o h`am thˆong thu.`o.ng cu’ a h`am mˆo. t biˆe´n x khi cˆo´ di.nh gi´a tri.
cu’ a biˆe´n y. Do d´o c´ac da. o h`am riˆeng du.o.
.c t´ınh theo c´ac quy t˘a´c v`a
cˆong th´u.c t´ınh da. o h`am thˆong thu.`o.ng cu’ a h`am mˆo. t biˆe´n. . ta c´o thˆe’ di.nh ngh˜ıa da.o h`am riˆeng Nhˆa. n x´et. Ho`an to`an tu.o.ng tu. cu’ a h`am ba (ho˘a. c nhiˆe`u ho.n ba) biˆe´n sˆo´. .p
9.1.2 D- a. o h`am cu’ a h`am ho.
Nˆe´u h`am w = f (x, y), x = x(t), y = y(t) th`ı biˆe’u th´u.c w =
f [x(t), y(t)] l`a h`am ho. .p cu’ a t. Khi d´o · · · = + (9.1) dw
dt ∂w
∂x dx
dt ∂w
∂y dy
dt Nˆe´u w = f (x, y), trong d´o x = x(u, v), y = y(u, v) th`ı 9.1.3 H`am kha’ vi
Gia’ su.’ h`am w = f (M) x´ac di.nh trong mˆo. t lˆan cˆa. n n`ao d´o cu’ a diˆe’m
.c go. i l`a h`am kha’ vi ta. i diˆe’m M(x, y) nˆe´u sˆo´ gia
M(x, y). H`am f du.o. = + ,
(9.2) · = + ∂w
∂u
∂w
∂v ∂w
∂x
∂w
∂x ∂x
∂u
∂x
∂v ∂w
∂y
∂w
∂y ∂y
∂u
∂y
∂v 112 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n ∆f (M) = f (x + ∆, y + ∆y) − f (x, y) cu’ a h`am khi chuyˆe’n t`u. diˆe’m
M(x, y) dˆe´n diˆe’N (x + ∆, y + ∆y) c´o thˆe’ biˆe’u diˆe˜n du.´o.i da. ng ρ → 0 ∆f (M) = D1∆x + D2∆y + o(ρ), trong d´o ρ = p∆x2 + ∆y2. Nˆe´u h`am f (x, y) kha’ vi ta. i diˆe’m M(x, y) th`ı (M) = D1, (M) = D2 ∂f
∂x ∂f
∂y v`a khi d´o 9.1.4 D- a. o h`am theo hu.´o.ng
Gia’ su.’ : ∆f (M) = (M)∆x + ∆y + o(ρ), ρ → 0. (9.3) ∂f
∂x ∂f
∂y (1) w = f (M) l`a h`am x´ac di.nh trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m M(x, y); (2) ~e = (cos α, cos β) l`a vecto. do.n vi. trˆen du.`o.ng th˘a’ ng c´o hu.´o.ng L qua diˆe’m M(x, y); (3) N = N (x + ∆x, y + ∆y) l`a diˆe’m thuˆo. c L v`a ∆e l`a dˆo. d`ai cu’ a doa. n th˘a’ ng MN . Nˆe´u tˆo` n ta. i gi´o.i ha. n h˜u.u ha. n ∆w
∆` lim
∆`→0
(N →M ) .c go. i l`a da. o h`am ta. i diˆe’m M(x, y) theo hu.´o.ng cu’ a , t´u.c l`a th`ı gi´o.i ha. n d´o du.o.
vecto. ~e v`a du.o. .c k´y hiˆe.u l`a ∂w
∂~e · = lim
∆`→0 ∂w
∂~e ∆w
∆` 113 9.1. D- a. o h`am riˆeng .c t´ınh theo Da. o h`am theo hu.´o.ng cu’ a vecto. ~e = (cos α, cos β) du.o. cˆong th´u.c = (M) cos α + (M) cos β. (9.4) ∂f
∂~e ∂f
∂x ∂f
∂y trong d´o cos α v`a cos β l`a c´ac cosin chı’ phu.o.ng cu’ a vecto. ~e. , (t´u.c l`a vecto. (cid:16) Vecto. v´o.i c´ac to. a dˆo. ∂F
∂y ∂f
∂x ∂f
∂x (cid:17)) du.o.
.c go. i
.c k´y hiˆe.u l`a ∂f
v`a
∂y
l`a vecto. gradiˆen cu’ a h`am f (M) ta. i diˆe’m M(x, y) v`a du.o.
gradf (M). c´o biˆe’u th´u.c l`a T`u. d´o da. o h`am theo hu.´o.ng ∂f
∂~e = (cid:10)gradf, ~e(cid:11). ∂f
∂~e Ta lu.u ´y r˘a`ng: 1) Nˆe´u h`am w = f (x, y) kha’ vi ta. i diˆe’m M(x, y) th`ı n´o liˆen tu. c ta. i M v`a c´o c´ac da. o h`am riˆeng cˆa´p 1 ta. i d´o; 2) N´eu h`am w = f (x, y) c´o c´ac da. o h`am riˆeng cˆa´p 1 theo mo. i biˆe´n
trong lˆan cˆa. n n`ao d´o cu’ a diˆe’m M(x, y) v`a c´ac da. o h`am riˆeng n`ay liˆen
tu. c ta. i diˆe’m M(x, y) th`ı n´o kha’ vi ta. i diˆe’m M. Nˆe´u h`am f (x, y) kha’ vi ta. i diˆe’m M(x, y) th`ı n´o c´o da. o h`am theo mo. i hu.´o.ng ta. i diˆe’m d´o. 9.1.5 D- a. o h`am riˆeng cˆa´p cao
Gia’ su.’ miˆe`n D ⊂ R2 v`a Ch´u ´y. Nˆe´u h`am f (x, y) c´o da.o h`am theo mo. i hu.´o.ng ta. i diˆe’m M0
th`ı khˆong c´o g`ı da’ m ba’ o l`a h`am f (x, y) kha’ vi ta. i diˆe’m M0 (xem v´ı
du. 4). f : D → R 114 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n l`a h`am hai biˆe´n f (x, y) du.o. .c cho trˆen D. Ta d˘a. t 6= ±∞o, Dx = n(x, y) ∈ D : ∃ 6= ±∞o. Dy = n(x, y) ∈ D : ∃ ∂f
∂x
∂f
∂y D∗ = Dx ∩ Dy v`a du.o. .c go. i l`a c´ac da. o ∂f
∂x ∂f
∂y D- i.nh ngh˜ıa. 1) C´ac da. o h`am riˆeng
h`am riˆeng cˆa´p 1.
2) Nˆe´u h`am : Dx → R v`a : Dy → R c´o c´ac da. o h`am riˆeng ∂f
∂x ∂f
∂y (cid:16) (cid:17) = = , ∂2f
∂x2 (cid:16) (cid:17) = , (cid:16) (cid:17) = , (cid:16) (cid:17) = = ∂
∂x
∂
∂y
∂
∂x
∂
∂y ∂f
∂x
∂f
∂x
∂f
∂y
∂f
∂y ∂2f
∂x∂x
∂2f
∂x∂y
∂2f
∂y∂x
∂2f
∂y∂y ∂2f
∂y2 th`ı ch´ung du.o. .c go. i l`a c´ac da. o h`am riˆeng cˆa´p 2 theo x v`a theo y. .c di.nh ngh˜ıa nhu. l`a c´ac da. o h`am riˆeng C´ac da. o h`am riˆeng cˆa´p 3 du.o.
cu’ a da. o h`am riˆeng cˆa´p 2, v.v... v`a .p
Ta lu.u ´y r˘a`ng nˆe´u h`am f (x, y) c´o c´ac da. o h`am hˆo˜n ho. ∂2f
∂x∂y .p n`ay liˆen tu. c ta. i diˆe’m (x, y) th`ı ta. i diˆe’m d´o c´ac da. o h`am hˆo˜n ho. ∂2f
∂y∂x
b˘a`ng nhau: · = ∂2f
∂x∂y ∂2f
∂y∂x C ´AC V´I DU. 115 9.1. D- a. o h`am riˆeng V´ı du. 1. T´ınh da.o h`am riˆeng cˆa´p 1 cu’ a c´ac h`am 1) 4w = x2 − 2xy2 + y3. du.o. Gia’ i. 1) Da. o h`am riˆeng 2) w = xy.
.c t´ınh nhu. l`a da. o h`am cu’ a h`am w ∂w
∂x theo biˆe´n x v´o.i gia’ thiˆe´t y = const. Do d´o x = 2x − 2y2 + 0 = 2(x − y2). = (x2 − 2xy2 + y3)0 ., ta c´o y = 0 − 4xy + 3y2 = y(3y − 4x). = (x2 − 2xy2 + y3)0 ∂w
∂x
Tu.o.ng tu.
∂w
∂y 2) Nhu. trong 1), xem y = const ta c´o 0
= (cid:0)xy(cid:1)
x = yxy−1. ∂w
∂x Tu.o.ng tu. = xylnx. .c
., khi xem x l`a h˘a`ng sˆo´ ta thu du.o.
∂w
∂y (v`ı w = xy l`a h`am m˜u dˆo´i v´o.i biˆe´n y khi x = const. N V´ı du. 2. Cho w = f (x, y) v`a x = ρ cos ϕ, y = ρ sin ϕ. H˜ay t´ınh ∂w
∂ρ . v`a ∂w
∂ϕ
Gia’ i. Dˆe’ ´ap du. ng cˆong th´u.c (9.2), ta lu.u ´y r˘a`ng w = f (x, y) = f (ρ cos ϕ, ρ sin ϕ) = F (ρ, ϕ). Do d´o theo (9.2) v`a biˆe’u th´u.c dˆo´i v´o.i x v`a y ta c´o = + = cos ϕ + sin ϕ ∂w
∂y = + = (−ρ sin ϕ) + (ρ cos ϕ) ∂w
∂ρ
∂w
∂ϕ ∂w
∂x
∂w
∂x ∂x
∂ρ
∂x
∂ϕ ∂w
∂y
∂w
∂y ∂y
∂ρ
∂y
∂ϕ ∂w
∂x
∂w
∂x = ρ(cid:16) − sin ϕ + cos ϕ(cid:17). N ∂w
∂x ∂w
∂y
∂w
∂y −→ 116 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n V´ı du. 3. T´ınh da. o h`am cu’ a h`am w = x2 + y2x ta. i diˆe’m M0(1, 2) theo
hu.´o.ng cu’ a vecto. M0M1, trong d´o M1 l`a diˆe’m v´o.i to. a dˆo. (3, 0). −→ −→ Gia’ i. Dˆa` u tiˆen ta t`ım vecto. do.n vi. ~e c´o hu.´o.ng l`a hu.´o.ng d˜a cho. Ta c´o √ ⇒ | = 2 ⇒ ~e = M0M1| = 2 M0M1 = (2, −2) = 2e1 − 2e2,
M0M1
|M0M1| 2e1 − 2e2
√
2 2 = ~e1 − ~e2. 1
√
2 1
√
2 trong d´o ~e1, ~e2 l`a vecto. do.n vi. cu’ a c´ac tru. c to. a dˆo. . T`u. d´o suy r˘a`ng · cos β = − , cos α = 1
√
2 1
√
2 x(1, 2) = 6, x = 2x + y2 ⇒ f 0
f 0
y = 2xy ⇒ f 0
f 0 x(M0) = f 0
y(M0) = f 0 y(1, 2) = 4. Tiˆe´p theo ta t´ınh c´ac da. o h`am riˆeng ta. i diˆe’m M0(1, 2). Ta c´o √ = 6 · 2. N − 4 · = .c
Do d´o theo cˆong th´u.c (9.4) ta thu du.o.
∂f
1
√
∂~e
2 1
√
2 V´ı du. 4. H`am f (x, y) = x + y + p|xy| c´o da. o h`am theo mo. i hu.´o.ng
ta. i diˆe’m O(0, 0) nhu.ng khˆong kha’ vi ta. i d´o. . tˆo` n ta. i da. o h`am theo mo. i hu.´o.ng. Gia’ i. 1. Su.
Ta x´et hu.´o.ng cu’ a vecto. ~e di ra t`u. O v`a lˆa. p v´o.i tru. c Ox g´oc α. Ta c´o ∆ef (0, 0) = ∆x + ∆y + p|∆x∆y| = (cid:0) cos α + sin α + p| cos α sin α|(cid:1)ρ, 117 9.1. D- a. o h`am riˆeng trong d´o ρ = p∆x2 + ∆y2, ∆x = ρ cos α, ∆y = ρ sin α. T`u. d´o suy ra = cos α + sin α + p| sin α cos α| (0, 0) = lim
ρ→0 ∂f
∂~e ∆ef (0, 0)
ρ
t´u.c l`a da. o h`am theo hu.´o.ng tˆo` n ta. i theo mo. i hu.´o.ng. 2. Tuy nhiˆen h`am d˜a cho khˆong kha’ vi ta. i O. Thˆa. t vˆa. y, ta c´o
∆f (0, 0) = f (∆x, ∆y) − f (0, 0) = ∆x + ∆y + p|∆x| |∆y| − 0. x = 1 v`a f 0
y = 1 (ta. i sao ? ) nˆen nˆe´u f kha’ vi ta. i O(0, 0) th`ı
∆f (0, 0) = ∆x + ∆y + p|∆x∆y| = 1 · ∆x + 1 · ∆y + ε(ρ)ρ V`ı f 0 ε(ρ) → 0(ρ → 0), ρ = p∆x2 + ∆y2 hay l`a lu.u ´y ∆x = ρ cos α, ∆y = ρ sin α ta c´o ε(ρ) = p| cos α sin α|.
Vˆe´ pha’ i d˘a’ ng th´u.c n`ay khˆong pha’ i l`a vˆo c`ung b´e khi ρ → 0 (v`ı n´o
ho`an to`an khˆong phu. thuˆo. c v`ao ρ). Do d´o theo di.nh ngh˜ıa h`am f (x, y)
d˜a cho khˆong kha’ vi ta. i diˆe’m O. N
V´ı du. 5. T´ınh c´ac da.o h`am riˆeng cˆa´p 2 cu’ a c´ac h`am: · 2) w = arctg x
1) w = xy,
y
Gia’ i. 1) Dˆa` u tiˆen t´ınh c´ac da. o h`am riˆeng cˆa´p 1. Ta c´o = yxy−1, = xylnx. ∂w
∂x ∂w
∂y = y(y − 1)xy−2, = xy−1 + yxy−1lnx = xy−1(1 + ylnx), = yxy−1lnx + xy · = xy−1(1 + ylnx), 1
x = xy(lnx)2. Tiˆe´p theo ta c´o
∂2w
∂x2
∂2w
∂y∂x
∂2w
∂x∂y
∂2f
∂y2 118 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n 2) Ta c´o · = = − , ∂w
∂x y
x2 + y2 ∂w
∂y x
x2 + y2 T`u. d´o (cid:16) = , (cid:17) = − (cid:16) (cid:17) = = , 2xy
(x2 + y2)2
2xy
x2 + y2 (cid:16) = , (cid:17) = x2 − y2
(x2 + y2)2 (cid:16) − · (cid:17) = = ∂2w
∂x2
∂2w
∂y2
∂2w
∂x∂y
∂2w
∂y∂x ∂
∂x
∂
∂y
∂
∂y
∂
∂x y
x2 + y2
−x
x2 + y2
y
x2 + y2
x
x2 + y2 x2 − y2
(x2 + y2)2 = . N Nhˆa. n x´et. Trong ca’ 1) lˆa˜n 2) ta dˆe`u c´o ∂2w
∂x∂y ∂2w
∂y∂x V´ı du. 6. T´ınh c´ac da. o h`am riˆeng cˆa´p 1 cu’ a h`am w = f (x + y2, y + x2)
ta. i diˆe’m M0(−1, 1), trong d´o x v`a y l`a biˆe´n dˆo. c lˆa. p. Gia’ i. D˘a. t t = x + y2, v = y + x2. Khi d´o w = f (x + y2, y + x2) = f (t, v). Nhu. vˆa. y w = f (t, v) l`a h`am ho.
.p cu’ a hai biˆe´n dˆo. c lˆa. p x v`a y. N´o phu.
thuˆo. c c´ac biˆe´n dˆo. c lˆa. p thˆong qua hai biˆe´n trung gian t, v. Theo cˆong
th´u.c (9.2) ta c´o: v(x + y2, y + x2) · 2x · · = + ∂w
∂x ∂v
∂x ∂t
∂x ∂f
∂f
∂t
∂v
t(x + y2, y + x2) · 1 + f 0
= f 0
t + 2xf 0
= f 0
v. 119 9.1. D- a. o h`am riˆeng t(0, 2) − 2f 0 v(0, 2) (−1, 1) = (0, 2) = f 0 ∂w
∂x t(·)2y + f 0 v(·)1 · · + = f 0 = ∂f
∂t ∂t
∂y ∂w
∂y ∂v
∂y ∂f
∂x
∂f
∂v t + f 0 v = 2yf 0 t(0, 2) + f 0 v(0, 2). N (−1, 1) = (0, 2) = 2f 0 ∂w
∂y ∂f
∂y B `AI T ˆA. P T´ınh da. o h`am riˆeng cu’ a c´ac h`am sau dˆay y = 3y2 + 9x2y2) (DS. f 0 1. f (x, y) = x2 + y3 + 3x2y3.
x = 2x + 6xy3, f 0 2. f (x, y, z) = xyz + . x = yz + y = xz − z = xy − , f 0 , f 0 ) (DS. f 0 x
yz
1
yz x
y2z x
yz2 x = y cos(xy + yz), y = (x + z) cos(xy + yz), f 0
f 0 3. f (x, y, z) = sin(xy + yz). (DS. f 0
z = y cos(xy + yz)) 4. f (x, y) = tg(x + y)ex/y. , (DS. f 0
x = + tg(x + y)ex/y 1
y + tg(x + y)ex/y(cid:16) − (cid:17).) f 0
y = ex/y
cos2(x + y)
ex/y
cos2(x + y) x
y2 x = y = 5. f = arc sin . (DS. f 0 , f 0 ) |y|
x2 + y2 −xsigny
x2 + y2 x
px2 + y2 x = yln(xy) + y, f 0 y = xln(xy) + x) 6. f (x, y) = xyln(xy). (DS. f 0 z 120 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n z−1 z (cid:17) 7. f (x, y, z) = (cid:16) . y
x z (cid:17) (cid:16) (cid:17) (cid:16) − , (DS. y
x (cid:17) (cid:17) ) , ln f 0
y = y
x2
f 0
z = (cid:16) f 0
x = z(cid:16)
z
y y
x
y
(cid:16)
x (cid:17) = −
y
z
x z
x
y
x 8. f (x, y, z) = zx/y. x = xx/ylnz · (cid:16) y = zx/ylnz · (cid:16) z = (cid:16) (cid:17), f 0 (DS. f 0 (cid:17)zx/y−1) (cid:17), f 0 −x
y2 x
y 1
y 9. f (x, y, z) = xyz . y = xyz zyz−1lnx, f 0 z = xyz ln(x)zlny) x = yzxyz−1, f 0
10. f (x, y, z) = xyyzzx. y = xylnxyzzx + xyyz−1zx+1, x = xy−1yz+1zx + xyyzzxlnz, f 0
(DS. f 0
z = xyyzlny · zx + xy+1yzzx−1)
f 0 (DS. f 0 . 11. f (x, y) = ln sin y = − x = , f 0 cotg ) cotg (DS. f 0 x + a
y x + a
√
y x + a
√
y
x + a
√
y 1
√
y 12. f (x, y) = − exarctgy. x = y = − − − exarctgy, f 0 ) (DS. f 0 x
y
1
y x
y2 ex
1 + y2 13. f (x, y) = ln(cid:0)x + px2 + y2(cid:1). x = y = · (DS. f 0 , f 0 ). 1
x + px2 + y2 y
px2 + y2
.p sau dˆay (gia’ thiˆe´t h`am f (x, y) 1
px2 + y2
T`ım da. o h`am riˆeng cu’ a h`am ho. kha’ vi) x = f 0 t + f 0 v2y, t = x + y, v = x2 + y2) (DS. f 0 14. f (x, y) = f (x + y, x2 + y2).
y = f 0 v2x, f 0
t + f 0
y
x
x
y , (cid:17). 15. f (x, y) = f (cid:16) 121 9.1. D- a. o h`am riˆeng x = v, f 0 y = t + , v = ) (DS. f 0 f 0
t − f 0
v, t = −x
y2 f 0 1
x x
y y
x y
1
x2 f 0
y
16. f (x, y) = f (x − y, xy). x = f 0 y = −f 0 v, f 0 t + xf 0 v, t = x − y, v = xy) t + yf 0
17. f (x, y) = f (x − y2, y − x2, xy). (DS. f 0 x = f 0 t − 2xf 0 v + yf 0 w, f 0 y = −2yf 0 t + f 0 v + xf 0
w, (DS. f 0 t = x − y2, v = y − x2, w = xy) √ z2 + x2). 18. f (x, y, z) = f (px2 + y2, py2 + z2, x = + , + , (DS. f 0 f 0
y = yf 0
v√
x2 + z2 yf 0
t
px2 + y2 + , t = px2 + y2, f 0
z = xf 0
t
px2 + y2
zf 0
v
px2 + y2 xf 0
w√
z2 + x2
zf 0
w√
z2 + x2
√ z2 + x2) v = py2 + z2, w = 19. w = f (x, xy, xyz). t + yf 0
xf 0 u + yzf 0
v,
u + xzf 0
v,
xyf 0
v x = f 0
f 0
f 0
y =
f 0
z =
t = x, u = xy, v = xyz). (DS. Trong c´ac b`ai to´an sau dˆay h˜ay ch´u.ng to’ r˘a`ng h`am f (x, y) tho’a m˜an phu.o.ng tr`ınh d˜a cho tu.o.ng ´u.ng (f (x, y)-kha’ vi). − x = 0. 20. f = f (x2 + y2), y + 2y = nf . (cid:17), x 21. f = xnf (cid:16) ∂f
∂x
∂f
∂y ∂f
∂y
∂f
∂y y
x2 + xy = xyf . ∂f
∂y 23. f = − xy + y2 = 0. 22. f = yf (x2 − y2), y2 ∂f
∂x
+ f (x, y), x2 ∂f
∂x y2
3x ∂f
∂y 122 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n 24. f = xnf (cid:16) , (cid:17), x + αy + βz = nf . y
xα z
xβ ∂f
∂y ∂f
∂z , lnx + xf (cid:16) + y + z = f + . (cid:17), x 25. f = xy
z y
x ∂f
∂x ∂f
∂y ∂f
∂z xy
z 26. T´ınh , , nˆe´u f = cos(xy) ∂f
∂x
z
x
∂2f
∂y2 ∂2f
∂x∂y xy = − sin xy − xy cos xy, f 00 yy = ∂2f
∂x2
xx = −y2 cos xy, f 00
f 00 xz = −y sin t, f 00 yy = −z2 sin t, (DS.
−x2 cos xy) xx = − sin t, f 00
(DS. f 00
yz = −yz sin t, f 00
f 00 27. T´ınh c´ac da. o h`am riˆeng cˆa´p hai cu’ a h`am f = sin(x + yz).
xy = −z sin t, f 00
zz = −y2 sin t, t = x + yz) 28. T´ınh nˆe´u f = px2 + y2ex+y. h − xy + (x + y)(x2 + y2) + (x2 + y2)2i) (DS. ∂2f
∂x∂y
ex+y
(x2 + y2)3/2 xz = xyz−1y(1 + yzlnx), 29. T´ınh , , nˆe´u f = xyz . ∂2f
∂y∂z ∂2f
∂x∂z ∂2f
∂x∂y
xy = xyz−1z(1 + yzlnx), f 00
(DS. f 00
f 00
yz = lnx · xyz(1 + yzlnx)) nˆe´u f = arctg = 0) . (DS. 30. T´ınh ∂2f
∂x∂y xy(0, 0), f 00 31. T´ınh f 00 ∂2f
∂x∂y
xx(0, 0), f 00 x + y
1 − xy
yy(0, 0) nˆe´u f (x, y) = (1 + x)m(1 + y)n. xy(0, 0) = mn, f 00 yy (0, 0) = n(n − 1)) (DS. f 00 z
(cid:17) (DS. ) 32. T´ınh nˆe´u r = px2 + y2 + z2. r2 − x2
r3 xx(0, 0) = m(m − 1), f 00
∂2r
∂x2
xy, f 00 xz nˆe´u f = (cid:16) yz , f 00 z−1 33. T´ınh f 00 . xy = −z2y−2(cid:0)xy−1(cid:1) xz = (cid:16) (cid:17)(cid:16) (cid:17) (cid:16)1 + zln (cid:17), x
y
z−1, f 00 (DS. f 00 1
y x
y x
y z 123 9.1. D- a. o h`am riˆeng (cid:16) (cid:17) (cid:17)) · (cid:16)1 + zln f 00
yz = − 1
y x
y nˆe´u f = arc sin = . 34. Ch´u.ng minh r˘a`ng rx − y
x ∂2f
∂x∂y x
y
∂2f
∂y∂x
T´ınh c´ac da. o h`am cˆa´p hai cu’ a c´ac h`am (gia’ thiˆe´t hai lˆa` n kha’ vi) tv + 4x2f 00 xx = f 00
(DS. u00
xy = f 00
u00
yy = f 00
u00 vv + 2f 0
tt + 4xf 00
v,
tv + 4xyf 00
tt + 2(x + y)f 00
vv,
tt + 4yf 00
vv + 2f 0
tv + 4y2f 00
v,
t = x + y, v = x2 + y2.) 35. u = f (x + y, x2 + y2). (cid:17). 36. u = f (cid:16)xy, x
y xx = y2f 00 tv + tt − xy = xyf 00
u00 (DS. u00 tt − 2 yy = x2f 00
u00 1
y2 f 00
vv,
vv + f 0
f 00
t − f 0
v, 1
y2 tt + 2f 00
x
y3
x2
y2 f 00
vv + f 0
v, x2
y4 2x
y3 ) t = xy, v = f 00
tv +
x
y xy = − sin y cos x · f 00, 37. u = f (sin x + cos y). (DS. u00
xx = cos2 x · f 00 − sin x · f 0, u00
yy = sin2 y · f 00 − cos y · f 0)
u00 4a2t 38. Ch´u.ng minh r˘a`ng h`am 1
√ e− (x−x0 )2 f = πt 2a
(trong d´o a, x0 l`a c´ac sˆo´) tho’a m˜an phu.o.ng tr`ınh truyˆe`n nhiˆe.t · ∂f
∂t = a2 ∂2f
∂x2 124 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n 39. Ch´u.ng minh r˘a`ng h`am f = trong d´o 1
r r = p(x − x0)2 + (y − y0)2 + (z − z0)2 tho’a m˜an phu.o.ng tr`ınh Laplace: + + = 0, r 6= 0. ∆f ≡ ∂2f
∂x2 ∂2f
∂y2 ∂2f
∂z2 Trong c´ac b`ai to´an 40 - 44 ch´u.ng minh r˘a`ng c´ac h`am d˜a cho tho’a
m˜an phu.o.ng tr`ınh tu.o.ng ´u.ng (gia’ thiˆe´t f v`a g l`a nh˜u.ng h`am hai lˆa` n
kha’ vi) 40. u = f (x − at) + g(x + at), + 41. u = xf (x + y) + yg(x + y), − 2 = 0. ∂2u
∂t2
∂2u
∂x2 = a2 ∂2u
∂x2
∂2u
∂x∂y (cid:17) + xg(cid:16) + 2xy = 0. 42. u = f (cid:16) y
x y
x (cid:17), x2 ∂2u
∂x2 ∂2u
∂x∂y ∂2u
∂y2
+ y2 ∂2u
∂y2 (cid:17), (cid:17) + x1−ng(cid:16) 43. u = xnf (cid:16) y
x + 2xy = n(n − 1)u. ∂2u
∂x∂y + y2 ∂2u
∂y2 y
x
x2 ∂2u
∂x2 · · · = 44. u = f (x + g(y)), ∂u
∂x ∂2u
∂x∂y ∂u
∂y ∂2u
∂x2 45. T`ım da. o h`am theo hu.´o.ng ϕ = 135◦ cu’ a h`am sˆo´ ) (DS. − f (x, y) = 3x4 + xy + y3 ta. i diˆe’m M(1, 2). √
2
2 (DS. 0) 46. T`ım da. o h`am cu’ a h`am f (x, y) = x3 − 3x2y + 3xy2 + 1 ta. i diˆe’m
M(3, 1) theo hu.´o.ng t`u. diˆe’m n`ay dˆe´n diˆe’m (6, 5). ) (DS. 47. T`ım da. o h`am cu’ a h`am f (x, y) = lnpx2 + y2 ta. i diˆe’m M(1, 1)
theo hu.´o.ng phˆan gi´ac cu’ a g´oc phˆa` n tu. th´u. nhˆa´t. √
2
2 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 125 ) 48. T`ım da. o h`am cu’ a h`am f (x, y, z) = z2 − 3xy + 5 ta. i diˆe’m
M(1, 2, −1) theo hu.´o.ng lˆa. p v´o.i c´ac tru. c to. a dˆo. nh˜u.ng g´oc b˘a`ng nhau.
(DS. − √
3
3 49. T`ım da. o h`am cu’ a h`am f (x, y, z) = ln(ex + ey + ez) ta. i gˆo´c to. a dˆo.
v`a hu.´o.ng lˆa. p v´o.i c´ac tru. c to. a dˆo. x, y, z c´ac g´oc tu.o.ng ´u.ng l`a α, β, γ. ) (DS. cos α + cos β + cos γ
3 √ ) (DS. 50. T´ınh da. o h`am cu’ a h`am f (x, y) = 2x2 − 3y2 ta. i diˆe’m M(1, 0) theo
hu.´o.ng lˆa. p v´o.i tru. c ho`anh g´oc b˘a`ng 120◦. (DS. −2)
51. T`ım da. o h`am cu’ a h`am z = x2 − y2 ta. i diˆe’m M0(1, 1) theo hu.´o.ng
vecto. ~e lˆa. p v´o.i hu.´o.ng du.o.ng tru. c ho`anh g´oc α = 60◦. (DS. 1 −
3)
52. T`ım da. o h`am cu’ a h`am z = ln(x2 + y2) ta. i diˆe’m M0(3, 4) theo
2
hu.´o.ng gradien cu’ a h`am d´o.
5
53. T`ım gi´a tri. v`a hu.´o.ng cu’ a vecto. gradien cu’ a h`am w = tgx − x + 3 sin y − sin3 y + z + cotgz (cid:17). , , ta. i diˆe’m M0(cid:16) π
3 π
2 , cos β = ) π
4
(DS. (gradw)M = ~i + 3
~j, cos α =
8 8
√
73 3
√
73 −→ 54. T`ım da. o h`am cu’ a h`am w = arc sin ta. i diˆe’m M0(1, 1, 1) z
px2 + y2 ) (DS. theo hu.´o.ng vecto. M0M, trong d´o M = (3, 2, 3). 1
6 Trong mu. c n`ay ta x´et vi phˆan cu’ a h`am nhiˆe`u biˆe´n m`a dˆe’ cho go. n ta
chı’ cˆa` n tr`ınh b`ay cho h`am hai biˆe´n l`a du’ . Tru.`o.ng ho.
.p sˆo´ biˆe´n l´o.n
..
.c tr`ınh b`ay ho`an to`an tu.o.ng tu.
ho.n hai du.o. 9.2.1 Vi phˆan cˆa´p 1
Gia’ su.’ h`am w = f (x, y) kha’ vi ta. i diˆe’m M(x, y), t´u.c l`a ta. i d´o sˆo´ gia
to`an phˆa` n cu’ a h`am c´o thˆe’ biˆe’u diˆe˜n du.´o.i da. ng 126 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n ∆f (M) = f (x + ∆x, y + ∆y) − f (x, y) (9.5) = D1∆x + D2∆y + o(ρ) trong d´o ρ = p∆x2 + ∆y2, D1 v`a D2 khˆong phu. thuˆo. c v`ao ∆x v`a
∆y. Khi d´o biˆe’u th´u.c (go. i l`a phˆa` n ch´ınh tuyˆe´n t´ınh dˆo´i v´o.i ∆x v`a ∆y
cu’ a sˆo´ gia ∆f ) D1∆x + D2∆y
.c go. i l`a vi phˆan (hay vi phˆan to`an phˆa` n ≡ hay vi phˆan th´u. nhˆa´t)
.c k´y hiˆe.u l`a df : du.o.
cu’ a h`am w = f (x, y) v`a du.o. df = D1∆x + D2∆y. , V`ı ∆x = dx, ∆y = dy v`a v`ı f (x, y) kha’ vi ta. i M nˆen D1 = ∂f
∂y v`a D2 = ∂f
∂y dx + dy (9.6) df = ∂f
∂x ∂f
∂y Nhu. vˆa. y, nˆe´u w = f (x, y) kha’ vi ta. i M(x, y) th`ı t`u. (9.5) v`a (9.6) ta c´o ∆f (M) = df (M) + o(ρ) hay ∆f (M) = df (M) + ε(ρ)ρ (9.7) 9.2.2 ´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung
Dˆo´i v´o.i ∆x v`a ∆y du’ b´e ta c´o thˆe’ thay xˆa´p xı’ sˆo´ gia ∆f (M) bo.’ i vi
phˆan df (M), t´u.c l`a trong d´o ε(ρ) → 0 khi ρ → 0. ∆f (M) ≈ df (M) 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 127 hay l`a f (x + ∆x, y + ∆y) ≈ f (x, y) + (9.8) (M)∆x + (M)∆y ∂f
∂x ∂f
∂y Cˆong th´u.c (9.8) l`a co. so.’ dˆe’ ´ap du. ng vi phˆan t´ınh gˆa` n d´ung. Dˆo´i 9.2.3 C´ac t´ınh chˆa´t cu’ a vi phˆan ..
v´o.i h`am c´o sˆo´ biˆe´n nhiˆe`u ho.n 2 ta c˜ung c´o cˆong th´u.c tu.o.ng tu. Dˆo´i v´o.i c´ac h`am kha’ vi f v`a g ta c´o: (cid:17) = (i) d(f ± g) = df ± dg;
(ii) d(fg) = f dg + gdf , d(αf ) = αdf, α ∈ R;
(iii) d(cid:16) , g 6= 0; gdf − fdg
g2 f
g (iv) Vi phˆan cˆa´p 1 cu’ a h`am hai biˆe´n f (x, y) bˆa´t biˆe´n vˆe` da. ng bˆa´t 9.2.4 Vi phˆan cˆa´p cao luˆa. n x v`a y l`a biˆe´n dˆo.c lˆa. p hay l`a h`am cu’ a c´ac biˆe´n dˆo. c lˆa. p kh´ac. .c biˆe’u diˆe˜n bo.’ i cˆong th´u.c Gia’ su.’ h`am w = f (x, y) kha’ vi trong miˆe`n D. Khi d´o vi phˆan cˆa´p 1
cu’ a n´o ta. i diˆe’m (x, y) ∈ D tu.o.ng ´u.ng v´o.i c´ac sˆo´ gia dx v`a dy cu’ a c´ac
biˆe´n dˆo. c lˆa.p du.o. dx + dy. (9.9) df = ∂f
∂x ∂f
∂y .’ dˆay, dx = ∆x, dy = ∆y l`a nh˜u.ng sˆo´ gia t`uy ´y cu’ a biˆe´n dˆo. c lˆa. p, d´o
O
l`a nh˜u.ng sˆo´ khˆong phu. thuˆo. c v`ao x v`a y. Nhu. vˆa. y, khi cˆo´ di.nh dx v`a
dy vi phˆan df l`a h`am cu’ a x v`a y. Theo di.nh ngh˜ıa: Vi phˆan th´u. hai d2f (hay vi phˆan cˆa´p 2) cu’ a
.c di.nh ngh˜ıa nhu. l`a vi phˆan cu’ a vi h`am f (x, y) ta. i diˆe’m M(x, y) du.o.
phˆan th´u. nhˆa´t ta. i diˆe’m M v´o.i c´ac diˆe`u kiˆe.n sau dˆay: (1) Vi phˆan df l`a h`am chı’ cu’ a c´ac biˆe´n dˆo. c lˆa. p x v`a y. 128 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n x v`a f 0 y du.o. (2) Sˆo´ gia cu’ a c´ac biˆe´n dˆo.c lˆa. p x v`a y xuˆa´t hiˆe.n khi t´ınh vi phˆan
.c xem l`a b˘a`ng sˆo´ gia dˆa` u tiˆen, t´u.c l`a b˘a`ng dx v`a dy. cu’ a f 0 T`u. d´o (M)dxdy + d2f (M) = dx2 + 2 (M)dy2 (9.10) ∂2f (M)
∂x2 ∂2f
∂x∂y ∂2f
∂y2 trong d´o dx2 = (dx)2, dy2 = (dy)2 v`a ta xem c´ac da.o h`am riˆeng hˆo˜n
ho. .p b˘a`ng nhau.
Mˆo. t c´ach h`ınh th´u.c d˘a’ ng th´u.c (9.10) c´o thˆe’ viˆe´t du.´o.i da. ng 2
dy(cid:17) dx + d2f = (cid:16) f (x, y) ∂
∂x ∂
∂y .c hiˆe.n ph´ep “b`ınh phu.o.ng” ta cˆa` n diˆe`n f (x, y) v`ao t´u.c l`a sau khi thu.
“ˆo trˆo´ng”. .
Tu.o.ng tu. 3
dy(cid:17) dx + d3f = (cid:16) f (x, y) ∂
∂y dx3 + 3 dx2dy + 3 dxdy2 + dy3, = ∂
∂x
∂3f
∂x3 ∂3f
∂x2∂y ∂3f
∂x∂y2 ∂3f
∂y3 dxn−k dyk. (9.11) dnf (x, y) = C k
n ∂nf
∂xn−k∂yk v.v... Mˆo. t c´ach quy na. p ta c´o
n
X
k=0 .p nˆe´u Trong tru.`o.ng ho. w = f (t, v), t = ϕ(x, y), v = ψ(x, y) th`ı dw = dt + dx (t´ınh bˆa´t biˆe´n vˆe` da. ng !) ∂f
∂t ∂f
∂v dtdy + d2t + d2v. (9.12) d2w = ∂2f
∂t2 dt2 + 2 ∂2f
∂t∂v ∂2f
∂v2 dv2 + ∂f
∂t ∂f
∂v 9.2.5 Cˆong th´u.c Taylor 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 129 Nˆe´u h`am f (x, y) l`a n + 1 lˆa` n kha’ vi trong ε-lˆan cˆa. n V cu’ a diˆe’m
M0(x0, y0) th`ı dˆo´i v´o.i diˆe’m bˆa´t k`y M(x, y) ∈ V ta c´o cˆong th´u.c Taylor x(x0, y0)(x − x0) + f 0 y(x0, y0)(y − y0)(cid:1) f (x, y) = f (x0, y0) + (cid:0)f 0 1
1! xx(x0, y0)(x − x0)2 + 2f 00 xy(x0, y0)(x − x0)(y − y0) + (cid:0)f 00 + · · · + (x − x0)n−i(y − y0)i C i
n 1
n! ∂nf (x0, y0)
∂xn−i∂yi + (9.13) (x − x0)n−i(y − y0), 1
(n + 1)! ∂n+1f (ξ, η)
∂xn−i∂yi 1
2!
+ f 00
yy(x0, y0)(y − y0)(cid:1)
m
X
i=0
n
X
i=0 trong d´o ξ = x0 + θ(x − x0), η = y0 + θ(y − y0), 0 < θ < 1.
hay l`a f (x, y) = f (x0, y0) + df (x0, y0) + d2f (x0, y0) + . . . 1
1! 1
2! + dnf (x0, y0) + Rn+1, 1
n! (9.14) = Pn(x, y) + Rn+1 trong d´o Pn(x, y) go. i l`a da th´u.c Taylor bˆa. c n cu’ a hai biˆe´n x v`a y,
Rn+1 l`a sˆo´ ha. ng du.. Nˆe´u d˘a. t ρ = p∆x2 + ∆y2 th`ı (9.14) c´o thˆe’ viˆe´t du.´o.i da. ng ρ → 0, f (x, y) = Pn(x, y) + 0(ρ), o.’ dˆay Rn+1 = o(ρ) l`a phˆa` n du. da. ng Peano. 130 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n 9.2.6 Vi phˆan cu’ a h`am ˆa’n
Theo di.nh ngh˜ıa: biˆe´n w du.o.
x, y, ..., t nˆe´u n´o du.o. .c go. i l`a h`am ˆa’n cu’ a c´ac biˆe´n dˆo. c lˆa. p .c cho bo.’ i phu.o.ng tr`ınh F (x, y, . . . , w) = 0 .c dˆo´i v´o.i w. khˆong gia’ i du.o. Dˆe’ t´ınh vi phˆan cu’ a h`am ˆa’n w ta lˆa´y vi phˆan ca’ hai vˆe´ cu’ a phu.o.ng
tr`ınh (xem nhu. dˆo` ng nhˆa´t th´u.c) rˆo` i t`u. d´o t`ım dw. Dˆe’ t´ınh d2w ta cˆa` n
lˆa´y vi phˆan cu’ a dw v´o.i lu.u ´y r˘a`ng dx v`a dy l`a h˘a`ng sˆo´, c`on dw l`a vi
phˆan cu’ a h`am. .c vi phˆan dw b˘a`ng c´ach t´ınh c´ac da.o h`am Ta c˜ung c´o thˆe’ thu du.o. riˆeng: x = − y = − , w0 , . . . w0 F 0
x(·)
w(·)
F 0 F 0
y(·)
w(·)
F 0 rˆo` i thˆe´ v`ao biˆe’u th´u.c dw = dx + dy + · · · + dt, v.v... ∂w
∂x ∂w
∂y ∂w
∂t C ´AC V´I DU. V´ı du. 1. T´ınh vi phˆan df nˆe´u 0
f 0
x = (cid:0)xy2(cid:1) y = (cid:0)xy2)0
f 0 y = 2xy. x = y2, 1) f (x, y) = xy2, 2) f (x, y) = px2 + y2.
Gia’ i. 1) Ta c´o Do d´o df (x, y) = y2dx + 2xydy. 2) Ta t´ınh c´ac da. o h`am riˆeng: · , f 0
x = f 0
y = x
px2 + y2 y
px2 + y2 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 131 Do d´o dx + dy = · N df = x
px2 + y2 y
px2 + y2 xdx + ydy
px2 + y2 V´ı du. 2. T´ınh df (M0) nˆe´u f (x, y, z) = ex2+y2 +z2 v`a M0 = M0(0, 1, 2). Gia’ i. Ta c´o (M)dx + (M)dy + (M)dz, M = M(x, y, z). df (M) = ∂f
∂y ∂f
∂z ∂f
∂x Ta t´ınh c´ac da.o h`am riˆeng = 2xex2+y2 +z2 ⇒ (v`ı x = 0) (M0) = 0, = 2yex2+y2 +z2 ⇒ (M0) = 2e5, = 2zex2+y2+z2 ⇒ (M0) = 4e5. ∂f
∂x
∂f
∂y
∂f
∂z ∂f
∂x
∂f
∂y
∂f
∂z T`u. d´o df (M0) = 2e5dy + 4e5dz. N V´ı du. 3. T´ınh dw ta. i diˆe’m M0(−1, 1) nˆe´u w = f (x + y2, y + x2). Gia’ i. C´ach 1. T´ınh c´ac da. o h`am riˆeng cu’ a h`am f (x, y) theo x v`a theo y rˆo` i ´ap du. ng cˆong th´u.c (9.9). T`u. v´ı du. 4, mu. c 9.1 ta c´o t(0, 2) − 2f 0 v(0, 2) (M0) = f 0 t(0, 2) + f 0 v(0, 2) (M0) = 2f 0 ∂f
∂x
∂f
∂y t = x + y2, v = y + x2 v`a do d´o t(0, 2) − 2f 0 v(0, 2)(cid:3)dx + 2(cid:2)2f 0 t(0, 2) + f 0 v(0, 2)(cid:3)dy. df (M0) = (cid:2)f 0 132 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n C´ach 2. ´Ap du. ng t´ınh bˆa´t biˆe´n vˆe` da. ng cu’ a vi phˆan cˆa´p 1.
Ta c´o t = x + y2 ⇒ dt = dx + 2ydy,
v = y + x2 ⇒ dv = 2xdx + dy. Do d´o (0, 2)dv (0, 2)dt + df (M0) = ∂f
∂v v(0, 2)[2xdx + dy]
t (0, 2) + f 0 v(0, 2)(cid:3)dx + (cid:2)2f 0 v(0, 2)(cid:3)dy. N ∂f
∂t
t(0, 2)[dx + 2ydy] + f 0
= f 0
t(0, 2) − 2f 0
= (cid:2)f 0 V´ı du. 4. 1) Cho h`am f (x, y) = xy. H˜ay t`ım vi phˆan cˆa´p hai cu’ a f
nˆe´u x v`a y l`a biˆe´n dˆo.c lˆa. p. 2) T`ım vi phˆan cˆa´p hai cu’ a h`am f (x + y, xy) nˆe´u x v`a y l`a biˆe´n dˆo. c lˆa. p. Gia’ i. 1) T`u. v´ı du. 2, 1) v`a cˆong th´u.c (9.10) ta c´o dxdy + d2f = ∂2f
∂x2 dx2 + 2 ∂2f
∂x∂y ∂2f
∂y2 dy2, trong d´o = y(y − 1)xy−2, = xy−1(1 + ylnx) ∂2f
∂x2
∂2f
∂y2 = xy(lnx)2,
∂2f
∂x∂y v`a do d´o d2f = y(y − 1)xy−2dx2 + xy−1(1 + ylnx)dxdy + xy(lnx)2dy2. 2) Ta viˆe´t h`am d˜a cho du.´o.i da. ng u = f (t, v), trong d´o t = x + y, v = xy. 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 133 1+ C´ach I. T´ınh c´ac da. o h`am riˆeng rˆo` i ´ap du. ng (9.10). Ta c´o: t(x + y, xy) + f 0 v(x + y, xy) · y, = f 0 t(x + y, xy) + f 0 v(x + y, xy) · x, = f 0 tvy + f 00 tvy + f 00 vvy2 = f 00 ∂f
∂x
∂f
∂y
∂2f
∂x2 tt + f 00
tt + 2yf 00 tv + y2f 00
vv, = f 00 tt + f 00 tvx + f 00 tvy + f 00 vvxy + f 0 v = f 00 ∂2f
∂x∂y tt + (x + y)f 00 tv + xyf 00 vv + f 0
v, tt + f 00 tvx + f 00 tvx + f 00 vvx2 = f 00 tt + 2xf 00 tv + x2f 00
vv. ∂2f
∂y2 = f 00
= f 00 tt + 2yf 00 vv + f 0 v)dxdy d2f = (f 00 Thˆe´ c´ac da. o h`am riˆeng t`ım du.o.
tv + y2f 00 .c
.c v`ao (9.10) ta thu du.o.
tv + xyf 00
tt + (x + y)f 00 tt + 2xf 00 tv + x2f 00 vv)dx2 + 2(f 00
vv)dy2. + (f 00 .c kˆe´t qua’ n`ay nˆe´u lu.u ´y r˘a`ng v´o.i 2+ C´ach II. Ta c´o thˆe’ thu du.o. t = x + y ⇒ dt = dx + dy v`a v = xy → dv = xdy + ydx v`a t`u. d´o d2t = d(dx + dy) = d2x + d2y = 0 (v`ı x v`a y l`a biˆe´n dˆo. c lˆa. p) v`a d2v = d(xdy + ydx) = dxdy + dxdy = 2dxdy. (dx + dy)(xdy + ydx) d2f = (dx + dy)2 + 2 ´Ap du. ng (9.12) ta c´o
∂2f
∂t2 ∂2f
∂t∂v (2dxdy) · 0 + (xdy + ydx)2 + + tv + x2f 00 vv(cid:1)dy2 = (cid:0)f 00 ∂2f
∂v2
tt + 2yf 00 ∂f
∂v
tt + 2xf 00 tv + y2f 00
tt + (x + y)f 00 vv + f 0 v(cid:1)dxdy. N + 2(cid:0)f 00 ∂f
∂t
vv(cid:1)dx2 + (cid:0)f 00
tv + xyf 00 134 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n V´ı du. 5. ´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung c´ac gi´a tri.: 1) a = (1, 04)2,03 − 1(cid:17) 2) b = arctg(cid:16) 1, 97
1, 02 3) c = p(1, 04)1,99 + ln(1, 02) . 4) d = sin 1, 49 · arctg0, 07
22,95 Gia’ i. Dˆe’ ´ap du. ng vi phˆan v`ao t´ınh gˆa` n d´ung ta cˆa` n thu. .c hiˆe.n c´ac bu.´o.c sau dˆay: Th´u. nhˆa´t l`a chı’ r˜o biˆe’u th´u.c gia’ i t´ıch dˆo´i v´o.i h`am m`a gi´a tri. gˆa` n d´ung cu’ a n´o cˆa` n pha’ i t´ınh. Th´u. hai l`a cho. n diˆe’m dˆa` u M0 sao cho gi´a tri. cu’ a h`am v`a cu’ a c´ac
da. o h`am riˆeng cu’ a n´o ta. i diˆe’m ˆa´y c´o thˆe’ t´ınh m`a khˆong cˆa` n d`ung
ba’ ng. x(x0, y0)∆x + f 0 y(x0, y0)∆y. Cuˆo´i c`ung ta ´ap du. ng cˆong th´u.c
f (x0 + ∆x, y0 + ∆y) = f (x0, y0) + f 0 1) T´ınh a = (1, 04)2,03. Ta x´et h`am f (x, y) = xy. Sˆo´ a cˆa` n t´ınh l`a gi´a tri. cu’ a h`am khi x = 1, 04 v`a y = 2, 03. Ta lˆa´y M0 = M0(1, 2). Khi d´o ∆x = 0, 04, ∆y = 0, 03.
Tiˆe´p theo ta c´o = yxy−1 ⇒ = 2 (cid:12)
(cid:12)M0 = xylnx ⇒ = 1 · ln1 = 0. ∂f
∂x
∂f
∂y ∂f
∂x
∂f
∂y (cid:12)
(cid:12)M0 Bˆay gi`o. ´ap du. ng cˆong th´u.c v`u.a nˆeu o.’ trˆen ta c´o: a = f (1, 04; 2, 03) = (1, 04)2,03 ≈ f (1, 2) + 2 · 0, 04 = 1 + 0, 08 = 1, 08. − 1(cid:17) l`a gi´a tri. cu’ a h`am 2) Ta nhˆa. n x´et r˘a`ng arctg(cid:16) 1, 97
1, 02 − 1(cid:17) f (x, y) = arctg(cid:16) x
y 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 135 ta. i diˆe’m M(1, 97; 1, 02). Ta cho.n M0 = M0(2, 1) v`a c´o ∆x = 1, 97 − 2 = −0, 03, ∆y = 1, 02 − 1 = 0, 02. Tiˆe´p dˆe´n ta c´o 2 = = ∂f
∂x y
y2 + (x − y)2 − 1(cid:17) 1 + (cid:16) 1
y
x
y · = − ∂f
∂y x
y2 + (x − y)2 T`u. d´o x(2, 1) = = 0, 5 (M0) = f 0 1
12 + (2 − 1)2 y(2, 1) = −1. (M0) = f 0 ∂f
∂x
∂f
∂y Do d´o arctg(cid:16) − 1(cid:17) = arctg(cid:16) − 1(cid:17) + (0, 5) · (−0, 03) + 1 · (0, 02) 1, 97
1, 02 2
1 = − 0, 015 − 0, 02 = 0, 785 − 0, 035 π
4
= 0, 75. √ xy + lnz ta. i diˆe’m M(1, 04; 1, 99; 1, 02). 3) Ta thˆa´y r˘a`ng c = p(1, 04)1,99 + ln(1, 02) l`a gi´a tri. cu’ a h`am
u = f (x, y, z) =
Ta cho.n M0 = M0(1, 2, 1). Khi d´o ∆x = 1, 04 − 1 = 0, 04 ∆y = 1, 99 − 2 = −0, 01 ∆z = 1, 02 − 1 = 0, 02. 136 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n ⇒ = = 1, (M0) = √
2 2 · 1
1 + ln1 √
2 ⇒ = (M0) = 0, √
2 yxy−1
xy + lnz
xylnx
xy + lnz
1 √ · ⇒ = (M0) = Bˆay gi`o. ta t´ınh gi´a tri. c´ac da.o h`am riˆeng ta. i diˆe’m M0. Ta c´o
∂f
∂x
∂f
∂y
∂f
∂z ∂f
∂x
∂f
∂y
∂f
∂z 1
2 xy + lnz 2z T`u. d´o suy ra √ 1 + ln1 + 1 · (0, 04) + 0 · (−0, 01) p(1, 04)1,99 + ln(1, 02) ≈ + (1/2) · 0, 02 = 1, 05. 4) Ta thˆa´y d l`a gi´a tri. cu’ a h`am f (x, y, z) = 2x sin y arctgx ta. i diˆe’m M(−2, 95; 1, 49; 0, 07) , 0(cid:17). Khi d´o Ta lˆa´y M0 = M0(cid:16) − 3, π
2 ∆x = −2, 95 − (−3) = 0, 05 ∆y = 1, 49 − 1, 57 = −0, 08 ∆z = 0, 07. Tiˆe´p theo ta c´o = 0, (cid:12)M0
= 0, (cid:12)M0
= 2−3. f 0
z(M0) = f (M0) = 2−3 sin(π/2) arctg0 = 0,
f 0
x(M0) = 2xln2 · sin y arctgz(cid:12)
f 0
y(M0) = 2x cos y arctgz(cid:12)
2x sin y
1 + z2 (cid:12) (cid:12)M0 .c
T`u. d´o ta thu du.o. ≈ 2−3 · 0, 07 ≈ 0, 01. N sin 1, 49 arctg0, 07
22,95 V´ı du. 6. Khai triˆe’n h`am f (x, y) = xy theo cˆong th´u.c Taylor ta. i lˆan
cˆa. n diˆe’m (1, 1) v´o.i n = 3. 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 137 Gia’ i. Trong tru.`o.ng ho. + + (*) f (x, y) = f (1, 1) + + R3. .p n`ay cˆong th´u.c Taylor c´o da. ng sau dˆay
df (1, 1)
1! d2f (1, 1)
2! d2f (1, 1)
3! 1+ T´ınh mo. i da. o h`am riˆeng cu’ a h`am cho dˆe´n xˆa´p 3. Ta c´o f 0
y = xylnx, f (3)
x2y = (2y − 1)xy−2 + y(y − 1)xy−2lnx, f 00
x2 = y(y − 1)xy−2,
f 0
x = yxy−1,
f 00
xy = xy−1 + yxy−1lnx,
f 00
y2 = xy(lnx)2,
f (3)
x3 = y(y − 1)(y − 2)xy−3,
f (3)
xy2 = 2xy−1lnx + yxy−1(lnx)2, f (3)
y3 = xy(lnx)3. 2+ T´ınh gi´a tri. cu’ a c´ac da. o h`am riˆeng ta. i diˆe’m (1, 1). Ta c´o
f 0
y(1, 1) = 0,
f (3)
x3 (1, 1) = 0, f 00
x2(1, 1) = 0,
f (3)
x2y(1, 1) = 1, f (1, 1) = 1,
f 00
xy(1, 1) = 1,
f (3)
xy2 (1, 1) = 0, y(1, 1)∆y = ∆x, x(1, 1)∆x + f 0
x2(1, 1)∆x2 + 2f 00 xy(1, 1)∆x∆y + f 00 y2 (1, 1)∆y2 = 2∆x∆y, f 0
x(1, 1) = 1,
f 00
y2 (1, 1) = 0,
f (3)
y3 (1, 1) = 0.
3+ Thˆe´ v`ao cˆong th´u.c (*) ta c´o df (1, 1) = f 0
d2f (1, 1) = f 00
d3f (1, 1) = 3∆x2∆y v`a do d´o xy = 1 + ∆x + ∆x∆y + ∆x2∆y + R3. N 1
2 .c cho bo.’ i phu.o.ng V´ı du. 7. T´ınh vi phˆan cu’ a h`am ˆa’n w(x, y) du.o.
tr`ınh w3 + 3x2y + xw + y2w2 + y − 2x = 0.
Gia’ i. Ta xem phu.o.ng tr`ınh d˜a cho nhu. mˆo. t dˆo` ng nhˆa´t v`a lˆa´y vi phˆan cu’ a vˆe´ tr´ai v`a vˆe´ pha’ i: 3w2dw + 6xydx + 3x2dy + wdx + xdw + 2y · w2dy + 2y2wdw − 2dx + dy = 0 138 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n v`a t`u. d´o r´ut ra dw. Ta c´o (6xy + w − 2)dx + (3x2 + 2yw2 + 1)dy + (3w2 + x + 2y2w)dw = 0 v`a do d´o dw = dx − dy. N 2 − 6xy − w
3w2 + x + 2y2w 3x2 + 2yw2 + 1
3w2 + x + 2y2w .c cho bo.’ i phu.o.ng V´ı du. 8. T´ınh dw v`a d2w cu’ a h`am ˆa’n w(x, y) du.o.
tr`ınh = 1. + + x2
2 y2
6 w2
8
. nhu. trong v´ı du. 7 ta c´o
Gia’ i. Dˆa` u tiˆen t`ım dw. Tu.o.ng tu. + = 0 ⇒ dw = − dx − dy. (*) xdx + ydy
3 wdw
4 4y
3w .c v´o.i lu.u ´y l`a dx, dy l`a 4x
w
La. i lˆa´y vi phˆan to`an phˆa` n d˘a’ ng th´u.c thu du.o.
h˘a`ng sˆo´; dw l`a vi phˆan cu’ a h`am. Ta c´o · dx − dy d2w = −4 wdx − xdw
w2 4
3 wdy − ydw
w2 hay l`a dx2 − dy2 − d2w = 4(cid:16) dxdw + dydw(cid:17) (**) 1
w x2
w2 1
3w y
3w2 .c x´ac di.nh bo.’ i hˆe. Dˆe’ c´o biˆe’u th´u.c d2w qua x, y, w, dx v`a dy ta cˆa` n thˆe´ dw t`u. (*) v`ao
(**). N
V´ı du. 9. C´ac h`am ˆa’n u(x, y) v`a v(x, y) du.o. xy + uv = 1, xv − yu = 3. 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 139 T´ınh du(1, −1), d2u(1, −1); dv(1, −1), d2v(1, −1) nˆe´u u(1, −1) = 1,
v(1, −1) = 2. Gia’ i. Lˆa´y vi phˆan hˆe. d˜a cho hai lˆa` n ta c´o (I) (ydx + xdy + udv + vdu = 0,
xdv + vdx − ydu − udy = 0. (II) (2dxdy + 2dudv + ud2v + vd2u = 0,
2dxdv − 2dudv + xd2v − yd2u = 0. Thˆe´ v`ao (I) gi´a tri. x = 1, y = −1, u = 1, v = 2 ta c´o ⇒ (III) (−dx + dy + dv + 2du = 0
2dx − dy + dv + du = 0 du = 3dx − 2dy
dv = −5dx + 3dy x = 3, u0 v = −2; v0 x = −5, v0 y = 3. .c u0 T`u. (III) ta c˜ung thu du.o. Thay v`ao (II) c´ac gi´a tri. x = 1, y = −1, u = 1, v = 2 v`a du, dv t`u. (III) ta c´o: d2v + 2d2u = −2dxdy − 2(3dx − 2dy)(3dy − 5dx)
d2v + d2u = 2dy(3dx − 2dy) − 2dx(3dy − 5dx) v`a do d´o d2u = 4(5dx2 − 10dxdy + 4dy2),
d2v = 10(−dx2 + 4dxdy − 2dy2). N B `AI T ˆA. P T´ınh vi phˆan dw cu’ a c´ac h`am sau (DS. dw = (2xy − y2)dx + (x2 − 2xy)dy) 1. w = x2y − y2x + 3. (DS. 6(x2 + y2)2(xdx + ydy)) 2. w = (x2 + y2)3. 3. w = x − 3 sin y. (DS. dw = dx − 3 cos ydy) 140 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n x−1 x x 4. w = ln(x2 + y). (DS. + ) 2xdx
x2 + y dy
x2 + y x
(cid:17) (cid:17) (cid:17) (cid:17) − (cid:16) idx + (cid:16) (DS. h(cid:16) dy) ln . 5. w = (cid:16) y
x y
x y
x y
x y
x
2ydx 2dy . 6. w = ln tg (DS. − + ). y
x x2 sin x sin 2y
x 2y
x T´ınh dw(M0) cu’ a c´ac h`am ta. i diˆe’m M0 d˜a cho (7-14) x , M0(1, 0). (DS. dw(1, 0) = −dy) dx + dy) (DS. dw(1, 1) = 7. w = e− y
√
8. w = y 3 x, M0(1, 1). 1
3 = −6dx + 3dy + 2dz) 9. f (x, y) = , M0(1, 2, 3). (DS. df (cid:12) yz
x (cid:17). , 10. f (x, y, z) = cos(xy + xz), M0(cid:16)1, (cid:12)M0
π
6 π
6 (cid:16) dx + dy + dz(cid:17)) = − (DS. df (cid:12) √
3
2 π
3 = 0) (cid:12)M0
11. f (x, y) = exy, M0(0, 0). (DS. df (cid:12) (cid:12)M0 = 12dx + 8ln2dy) 12. f (x, y) = xy, M0(2, 3). (DS. df (cid:12) = dx + dy) 13. f (x, y) = xln(xy), M0(1, 1). (cid:12)M0
(DS. df (cid:12) (cid:12)M0 14. f (x, y) = arctg (2dx − dy)). = (DS. df (cid:12) , M)(1, 2). x
y (cid:12)M0 1
5
.p sau dˆay ta. i c´ac diˆe’m d˜a chı’ ra (15-18) T`ım vi phˆan cu’ a c´ac h`am ho. 15. f (x, y) = f (x − y, x + y), M(x, y), M0(1, −1). v)dx + (f 0 t + f 0
t(2, 0) + f 0 v − f 0
t)dy,
v(2, 0)(cid:3)dx + (cid:2)f 0 v(2, 0) − f 0 t(2, 0)(cid:3)dy, (DS. df (cid:12) (cid:12)M df (cid:12) = (f 0
= (cid:2)f 0 (cid:12)M0 t = x − y, v = x + y) 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 141 16. f (x, y) = f (cid:16)xy, (cid:17), M(x, y), M0(0, 1). x
y t + t − (cid:17)dx + (cid:16)xf 0 (cid:17)dy, (DS. df (cid:12) f 0
v f 0
v x
y2 v(0, 0)(cid:3)dx, t = xy, v = ) df (cid:12) (cid:12)M = (cid:16)yf 0
= (cid:2)f 0 1
y
t(0, 0) + f 0 x
y (cid:12)M0 17. f (x, y, z) = f (x2 − y2, y2 − z2, z2 − x2), M(x, y, z), M0(1, 1, 1). w)dx + 2y(f 0 t)dy + 2z(f 0 v)dz, t − xf 0
t(0, 0, 0) − f 0 w − f 0
v(0, 0, 0) − f 0 t(0, 0, 0))dy v − f 0
w(0, 0, 0))dx + 2(f 0
v(0, 0, 0))dz, w(0, 0, 0) − f 0 (DS. df (cid:12) (cid:12)M = 2(xf 0
= 2(f 0 df (cid:12) (cid:12)M0 + 2(f 0
t = x2 − y2, v = y2 − z2, w = z2 − x2) 18. f (x, y, z) = f (sin x + sin y, cos x − cos z), M(x, y, z) v`a M0(0, 0, 0). t cos ydy + f 0 v sin zdz, (DS. df (cid:12) t cos x − f 0
t(0, 0)dx + f 0 v sin x)dx + f 0
v(0, 0)dy,
t = sin x + sin y, v = cos x − cos z). df (cid:12) (cid:12)M = (f 0
= f 0
(cid:12)M0 T´ınh vi phˆan dw v`a d2w ta. i diˆe’m M(x, y) (19-22) nˆe´u: 19. w = f (lnz), z = x2 + y2. tt − x2f 0 t + y2f 0 t)dx2 h(2x2f 00 (DS. d2w = tt − 4xyf 0 t)dxdy + (x2f 0 t − yf 0 t + 2yf 00 t2)dy2) 2
(x2 + y2)2
+ (4xyf 00 βdy + cf 0 20. w = f (α, β, γ), α = ax, β = by, γ = cz; a, b, c-h˘a`ng sˆo´. αdx + bf 0
α2dx2 + b2f 00 γdz;
β2dy2 + c2f 00 γ2dz2 (DS. dw = af 0
d2w = a2f 00 αβabdxdy + f 00 βγbcdydz + f 00 αγacdxdz)) + 2(f 00 21. w = f (x + y, x − y). (DS. x + y = u, x − y = v; uv + f 00 uv + f 00 u2 + 2f 00 v2)dx2 + (f 00 u2 − 2f 00 v2 )dxdy + (f 00 u2 − 2f 00 v2 )dy2) d2w = (f 00 142 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n f 0(cid:17)dx − (cid:17). 22. w = xf (cid:16) (DS. dw = (cid:16)f + f 0dy, x
y x
y x2
y2 d2w = (cid:16) f 0 + f 00)dx2 − (cid:16) f 0 + f 00(cid:17)dxdy − (cid:16) f 0 − f 00(cid:17)dy2) 2
y x
y2 2x2
y3 4x
y2 2x2
y3 x3
y4 T´ınh vi phˆan cˆa´p hai cu’ a c´ac h`am sau dˆay ta. i c´ac diˆe’m M(x, y)
v`a M0(x0, y0) nˆe´u f l`a h`am hai lˆa` n kha’ vi v`a x, y, z l`a biˆe´n dˆo. c lˆa. p
(23-25) tv(dx2 − dy2) + f 00 vv(dx + dy)2, 23. u = f (x − y, x + y), M(x, y), M0(1, 1) . tv(0, 2)(dx2 − dy2) tt(dx − dy)2 + 2f 00
tt(0, 2)dx(dx − dy)2 + 2f 00
+ f 00 vv(0, 2)(dx + dy)2) (DS. d2u(cid:12)
d2u(cid:12) (cid:12)M = f 00
= f 00
(cid:12)M0 24. u = f (x + y, z2), M(x, y, z), M0(−1, −1, 0). = f 00 (DS. d2u(cid:12) (cid:12)M v(0, 0)dz2, tt(dx + dy)2 + 4zf 00
tvdz(dx + dy)
vvdz2 + 2f 0
+ 4z2f 00
vd2z,
tt(0, 0)(dx + dy)2 + 2f 0
t = x + y, v = z2) = f 00 d2u(cid:12) (cid:12)M0 25. u = f (xy, x2 + y2), M(x, y), M0(0, 0). tv(ydz + xdy)(xdx + ydy) tt(ydx + xdy)2 + 4f 00
+ 4f 00 vv (xdx + ydy)2 + 2f 0 v(dx2 + dy2), (DS. d2u(cid:12) (cid:12)M = f 00 tdxdy + 2f 0
v(0, 0)(dx2 + dy2), t (0, 0)dxdy + 2f 0
t = xy, v = x2 + y2) = 2f 0 d2u(cid:12) (cid:12)M0 T´ınh vi phˆan dnw (26-27) nˆe´u: 26. w = f (ax + by + cz). (DS. dnw = f (n)(ax + by + cz)(adx + bdy + cdz)n) 9.2. Vi phˆan cu’ a h`am nhiˆe`u biˆe´n 143 n 27. w = f (ax, by, cz). dx + b dy + c dz(cid:17) (DS. dnw = (cid:16)a f (α, β, γ), ∂
∂α ∂
∂β ∂
∂γ α = ax, β = by, γ = cz) Khai triˆe’n c´ac h`am d˜a cho theo cˆong th´u.c Taylor dˆe´n c´ac sˆo´ ha. ng 28. f (x, y) = cˆa´p 2 (28-30) nˆe´u
1
x − y (DS. ∆w = + + R2) ∆y − ∆x
(x − y)2 ∆x2 − 2∆x∆y + ∆y2
(x − y)3 √ 29. f (x, y) = x + y. − (DS. ∆w = + R2) ∆x + ∆y
√
x + y
2 ∆x2 + 2∆x∆y + ∆y2
8(x + y)3/2 30. f (x, y) = ex+y. DS. ∆w = ex+y(∆x + ∆y) + ex+y (∆x + ∆y)2 + R2). 2 ´Ap du. ng vi phˆan dˆe’ t´ınh gˆa` n d´ung (31-35) (DS. ≈ 0, 94) 31. i) a = (0, 97)2,02 (DS. ≈ 4.998) ii) b = p(4, 05)2 + (2, 93)2 (DS. 1,05) 32. i) a = p(1.04)2,99 + ln 1, 02. √ xy + ln z. (DS. 1,013) Chı’ dˆa˜n. X´et h`am
ii) b = 3p(1, 02)2 + (0, 05)2.
Chı’ dˆa˜n. X´et h`am 3px2 + y2. 33. (DS. ≈ 0, 502)
(DS. ≈ 0, 273) i) a = sin 29◦ · tg46◦.
ii) b = sin 32◦ · cos 59◦. (DS. ≈ −0, 03) 34. i) a = ln(0, 093 + 0, 993). 144 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n Chı’ dˆa˜n. X´et h`am f = ln(x3 + y3), M0(0, 1).
(DS. ≈ 3, 037)
ii) b = p5e0,02 + (2, 03)2. Chı’ dˆa˜n. X´et h`am f = p5ex + y2, M0(0, 2). (DS. ≈ 2, 95) .
35. T´ınh vi phˆan cu’ a h`am f (x, y) = px3 + y3. ´U
ng du. ng dˆe’ t´ınh
xˆa´p xı’ p(1, 02)3 + (1, 97)3. Trong c´ac b`ai to´an sau dˆay (36-38) h˜ay t´ınh vi phˆan cˆa´p 1 cu’ a ) (DS. dz = 36. z3 + 3x2z = 2xy. h`am ˆa’n z(x, y) x´ac di.nh bo.’ i c´ac phu.o.ng tr`ınh tu.o.ng ´u.ng
(2y − 6xz)dx + 2xdy
3(x2 + z2) 37. cos2 x + cos2 y + cos2 z = 1. ). (DS. dz = − sin 2xdx + sin 2ydy
sin 2z (DS. dz = −dx − dy) 38. x + y + z = e−(x+y+z) . 39. Cho w l`a h`am cu’ a x v`a y x´ac di.nh bo.’ i phu.o.ng tr`ınh = ln + 1. x
w w
y T´ınh vi phˆan dw, d2w. , d2w = − ). (DS. dw = w(ydx + wdy)
y(x + w) w2(ydx − xdy)2
y2(x + w)2 .c x´ac di.nh bo.’ i phu.o.ng tr`ınh 40. T´ınh dw v`a d2w nˆe´u h`am w(x, y) du.o.
w − x = arctg
. y
w − x , (DS. dw = dx + (w − x)dy
(w − x)2 + y2 + y dy2). d2w = − 2(y + 1)(w − x)[(w − x)2 + y2]
[(w − x)2 + y2 + y]3 145 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n 9.3.1 Cu. .c da. i di.a phu.o.ng (ho˘a. c cu. f (M) < f (M0) (tu.o.ng ´u.ng : f (M) > f (M0)). .c da. i, cu. .c tiˆe’u cu’ a h`am sˆo´ l`a cu. .c tri. cu’ a h`am sˆo´. Go. i chung cu.
Diˆe`u kiˆe.n cˆa` n dˆe’ tˆo` n ta. i cu. .c tri.. .c tri. di.a phu.o.ng: Nˆe´u ta. i diˆe’m M0 h`am
.c tri. di.a phu.o.ng th`ı ta. i diˆe’m d´o ca’ hai da. o h`am riˆeng cˆa´p
f (x, y) c´o cu.
1 (nˆe´u ch´ung tˆo` n ta. i) dˆe`u b˘a`ng 0 ho˘a. c ´ıt nhˆa´t mˆo. t trong hai da. o h`am
riˆeng khˆong tˆo` n ta. i (d´o l`a nh˜u.ng diˆe’m t´o.i ha. n ho˘a. c diˆe’m d`u.ng cu’ a
h`am f (x, y)). Khˆong pha’ i mo. i diˆe’m d`u.ng dˆe`u l`a diˆe’m cu. Diˆe`u kiˆe.n du’ : gia’ su.’ f 00
xx(M0) =, f 00
xy(M0) = B, f 00
yy (M0) = C. Khi d´o: i) Nˆe´u ∆(M0) = > 0 v`a A > 0 th`ı ta. i diˆe’m M0 h`am f c´o (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) cu. (cid:12)
A B
(cid:12)
(cid:12)
B C
(cid:12)
(cid:12)
.c tiˆe’u di.a phu.o.ng. > 0 v`a A < 0 th`ı ta. i diˆe’m M0 h`am f c´o A B
B C (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) cu. ii) Nˆe´u ∆(M0) =
.c da. i di.a phu.o.ng. .a cu’ a f , t´u.c iii) Nˆe´u ∆(M0) = (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) iv) Nˆe´u ∆(M0) = = 0 th`ı M0 l`a diˆe’m nghi vˆa´n (h`am f c´o (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
A B
(cid:12)
< 0 th`ı M0 l`a diˆe’m yˆen ngu.
(cid:12)
B C
(cid:12)
(cid:12)
.c tri..
l`a ta. i M0 h`am f khˆong c´o cu.
(cid:12)
A B
(cid:12)
(cid:12)
B C
(cid:12)
(cid:12) .c tri. ta. i d´o). thˆe’ c´o v`a c˜ung c´o thˆe’ khˆong c´o cu. 146 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n .c tri. c´o diˆe`u kiˆe. n
.p do.n gia’ n nhˆa´t, cu.
.c tiˆe’u cu’ a h`am d´o da. t du.o. .c da.i ho˘a. c cu. 9.3.2 Cu.
Trong tru.`o.ng ho.
.c tri. c´o diˆe`u kiˆe.n cu’ a h`am f (x, y)
.c v´o.i diˆe`u kiˆe.n c´ac biˆe´n
l`a cu.
x v`a y tho’a m˜an phu.o.ng tr`ınh ϕ(x, y) = 0 (phu.o.ng tr`ınh r`ang buˆo. c).
.c tri. c´o diˆe`u kiˆe.n v´o.i diˆe`u kiˆe.n r`ang buˆo. c ϕ(x, y) ta lˆa. p Dˆe’ t`ım cu. .)
h`am Lagrange (h`am bˆo’ tro. F (x, y) = f (x, y)λϕ(x, y) .c tri. thˆong
.c x´ac di.nh v`a di t`ım cu.
. n`ay. Dˆay l`a phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh trong d´o λ l`a h˘a`ng sˆo´ nhˆan chu.a du.o.
thu.`o.ng cu’ a h`am bˆo’ tro.
Lagrange. T`ım diˆe`u kiˆe.n cˆa` n dˆe’ tˆo` n ta. i cu. .c tri. c´o diˆe`u kiˆe.n chung quy l`a gia’ i hˆe. phu.o.ng tr`ınh = + λ = 0 (9.15) = + λ = 0 ∂f
∂x
∂f
∂y ∂ϕ
∂x
∂ϕ
∂y ∂F
∂x
∂F
∂y ϕ(x, y) = 0
T`u. hˆe. n`ay ta c´o thˆe’ x´ac di.nh x, y v`a λ.
Vˆa´n dˆe` tˆo` n ta. i v`a d˘a. c t´ınh cu’ a cu. .c minh di.nh .c tri. di.a phu.o.ng du.o. .
trˆen co. so.’ x´et dˆa´u cu’ a vi phˆan cˆa´p hai cu’ a h`am bˆo’ tro. dxdy + d2F = dx2 + 2 dy2 ∂2F
∂x2 ∂2F
∂x∂y ∂2F
∂y2 .c t´ınh dˆo´i v´o.i c´ac gi´a tri. x, y, λ thu du.o. .c khi gia’ i hˆe. (9.15) v´o.i diˆe`u du.o.
kiˆe.n l`a dx + dy = 0 (dx2 + dy2 6= 0). ∂ϕ
∂x ∂ϕ
∂y Cu. thˆe’ l`a: 147 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n .c da. i c´o diˆe`u kiˆe.n.
.c tiˆe’u c´o diˆe`u kiˆe.n. .c tiˆe´n h`anh .c tri. cu’ a h`am ba biˆe´n ho˘a. c nhiˆe`u ho.n du.o. i) Nˆe´u d2F < 0 h`am f (x, y) c´o cu.
ii) Nˆe´u d2F > 0 h`am f (x, y) c´o cu.
iii) Nˆe´u d2F = 0 th`ı cˆa` n pha’ i kha’ o s´at.
Nhˆa. n x´et
i) Viˆe.c t`ım cu.
. nhu. o.’ 1. tu.o.ng tu. ii) Tu.o.ng tu. . c´o thˆe’ t`ım cu. .c tri. c´o diˆe`u kiˆe.n cu’ a h`am ba biˆe´n ho˘a. c
nhiˆe`u ho.n v´o.i mˆo. t ho˘a. c nhiˆe`u phu.o.ng tr`ınh r`ang buˆo. c (sˆo´ phu.o.ng
. v´o.i
tr`ınh r`ang buˆo. c pha’ i b´e ho.n sˆo´ biˆe´n). Khi d´o cˆa` n lˆa. p h`am bˆo’ tro.
sˆo´ th`u.a sˆo´ chu.a x´ac di.nh b˘a`ng sˆo´ phu.o.ng tr`ınh r`ang buˆo. c. iii) Ngo`ai phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh Lagrange, ngu.`o.i ta c`on 9.3.3 Gi´a tri. l´o.n nhˆa´t v`a b´e nhˆa´t cu’ a h`am
H`am kha’ vi trong miˆe`n d´ong bi. ch˘a. n da.t gi´a tri. l´o.n nhˆa´t (nho’ nhˆa´t)
ho˘a. c ta. i diˆe’m d`u.ng ho˘a. c ta. i diˆe’m biˆen cu’ a miˆe`n. d`ung phu.o.ng ph´ap khu.’ biˆe´n sˆo´ dˆe’ t`ım cu. .c tri. c´o diˆe`u kiˆe.n. C ´AC V´I DU.
.c tri. di.a phu.o.ng cu’ a h`am V´ı du. 1. T`ım cu. x v`a f 0 f (x, y) = x4 + y4 − 2x2 + 4xy − 2y2. y v`a t`ım c´ac diˆe’m t´o.i ha. n. Ta Gia’ i. i) Miˆe`n x´ac di.nh cu’ a h`am l`a to`an m˘a. t ph˘a’ ng R2.
ii) T´ınh c´ac da. o h`am riˆeng f 0 c´o f 0
x = 4x3 − 4x + 4y, f 0
y = 4y3 + 4x − 4y. Do d´o 4x3 − 4x + 4y = 0 4y3 + 4x − 4y = 0 148 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n x, f 0 v`a t`u. d´o √ √ 2 (x3 = 2
√ 2. (x1 = 0
y1 = 0 (x2 = −
√
2
y2 = Nhu. vˆa. y ta c´o ba diˆe’m t´o.i ha. n. V`ı f 0 y3 = −
y tˆo` n ta. i v´o.i mo. i diˆe’m M(x, y) ∈ R2 nˆen h`am khˆong c`on diˆe’m t´o.i ha. n n`ao kh´ac. iii) Ta t´ınh c´ac da. o h`am riˆeng cˆa´p hai v`a gi´a tri. cu’ a ch´ung ta. i c´ac diˆe’m t´o.i ha. n. f 00
xx(x, y) = 12x2 = 4, f 00
xy = 4, f 00
yy = 12y2 − 4. √
√ 2, +
2, − A = −4, B = 4, C = −4
2): A = 20, B = 4, C = 20
2): A = 20, B = 4, C = 20. Ta. i diˆe’m O(0, 0):
√
Ta. i diˆe’m M1(−
√
Ta. i diˆe’m M2(+
iv) Ta. i diˆe’m O(0, 0)ta c´o = = 16 − 16 = 0. (cid:12)
A B
(cid:12)
(cid:12)
B C
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
−4
4
(cid:12)
(cid:12)
4 −4
(cid:12)
(cid:12)
Dˆa´u hiˆe.u du’ khˆong cho ta cˆau tra’ l`o.i. Ta nhˆa. n x´et r˘a`ng trong lˆan
cˆa. n bˆa´t k`y cu’ a diˆe’m O tˆo` n ta. i nh˜u.ng diˆe’m m`a f (x, y) > 0 v`a nh˜u.ng
diˆe’m m`a f (x, y) < 0. Ch˘a’ ng ha. n do. c theo trung c Ox (y = 0) ta c´o = f (x, 0) = x4 − 2x2 = −x2(2 − x2) < 0 f (x, y)(cid:12) (cid:12)y=0 ta. i nh˜u.ng diˆe’m du’ gˆa` n (0, 0), v`a do. c theo du.`o.ng th˘a’ ng y = x f (x, y)(cid:12) (cid:12)y=x = f (x, x) = 2x4 > 0
Nhu. vˆa. y, ta. i nh˜u.ng diˆe’m kh´ac nhau cu’ a mˆo. t lˆan cˆa. n n`ao d´o cu’ a
diˆe’m O(0, 0) sˆo´ gia to`an phˆa` n ∆f (x, .y) khˆong c´o c`ung mˆo. t dˆa´u v`a do
d´o ta. i O(0, 0) h`am khˆong c´o cu. .c tri. di.a phu.o.ng. √ √ 2, 2) ta c´o Ta. i diˆe’m M1(− = = 400 − 16 > 0 A B
B C 20
4 4
20 (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) 149 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n √ √ 2, 2) h`am c´o cu. .c tiˆe’u di.a phu.o.ng v`a v`a A > 0 nˆen ta. i M1(−
fmin = −8. √ √ 2, − 2) ta c´o AC − B2 > 0 v`a A > 0 nˆen ta. i d´o Ta. i diˆe’m M2( h`am c´o cu. .c tiˆe’u di.a phu.o.ng v`a fmin = −8.
.c tri. cu’ a h`am V´ı du. 2. Kha’ o s´at v`a t`ım cu. f (x, y) = x2 + xy + y2 − 2x − 3y. Gia’ i. i) Hiˆe’n nhiˆen Df ≡ R.
ii) T`ım diˆe’m d`u.ng. Ta c´o ⇒ 2x + y − 2 = 0,
x + 2y − 3 = 0. f 0
x = 2x + y − 2
f 0
y = x + 2y − 3 , . Do d´o (cid:16) , y0 = .c c´o nghiˆe.m l`a x0 = 1
3 4
3 4
3 y tˆo` n tˆa. i ∀(x, y). xy = 1, C = f 00 1
Hˆe. thu du.o.
(cid:17) l`a diˆe’m
3
d`u.ng v`a ngo`ai diˆe’m d`u.ng d´o h`am f khˆong c´o diˆe’m d`u.ng n`ao kh´ac v`ı
x v`a f 0
f 0 x2 = 2, B f 00 y2 = 2. .c tri.. Ta c´o A = f 00 iii) Kha’ o s´at cu. Do d´o = 3 > 0 v`a A = 2 > 0 ∆(M0) = 2 1
1 2 (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12) , nˆen h`am f c´o cu. 4
3 1
3 (cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
.c tiˆe’u ta. i diˆe’m M0(
(cid:17). N
.c tri. cu’ a h`am f (x, y) = 6 − 4x − 3y v´o.i diˆe`u kiˆe.n l`a V´ı du. 3. T`ım cu.
x v`a y liˆen hˆe. v´o.i nhau bo.’ i phu.o.ng tr`ınh x2 + y2 = 1. Gia’ i. Ta lˆa. p h`am Lagrange F (x, y) = 6 − 4x − 3y + λ(x2 + y2 − 1). Ta c´o = −4 + 2λx, = −3 + 2λy ∂F
∂x ∂F
∂y 150 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n v`a ta gia’ i hˆe. phu.o.ng tr`ınh −4 + 2λx = 0 −3 + 2λx = 0 x2 + y2 = 1 Gia’ i ra ta c´o , , λ1 = x1 = y1 = 4
5 , , 5
2
λ2 = − x2 = − 3
5
y2 = − 5
2 3
5 4
5 V`ı = 0, = 2λ, = 2λ ∂2F
∂x2 ∂2F
∂x∂y ∂2F
∂y2 nˆen d2F = 2λ(dx2 + dy2). (cid:17) h`am , y = , x = , th`ı d2F > 0 nˆen ta. i diˆe’m (cid:16) 3
5 4
5 4
5 3
5 c´o cu. .c
th`ı d2F < 0 v`a do d´o h`am c´o cu. (cid:17). 5
Nˆe´u λ =
2
.c tiˆe’u c´o diˆe`u kiˆe.n.
5
4
Nˆe´u λ = −
, x = −
5
2
da. i c´o diˆe`u kiˆe.n ta. i diˆe’m (cid:16) − , y = −
4
5 3
5
3
, −
5 Nhu. vˆa. y = 11, + fmax = 6 + − = 1. N fmin = 6 − 16
5
16
5 9
5
9
5 V´ı du. 4. T`ım cu. x + y = 4. .c tri. c´o diˆe`u kiˆe.n cu’ a h`am
1) f (x, y) = x2 + y2 + xy − 5x − 4y + 10,
2) u = f (x, y, z) = x + y + z2 ( z − x = 1,
y − xz = 1. 151 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n Gia’ i. 1) T`u. phu.o.ng tr`ınh r`ang buˆo. c x + y = 4 ta c´o y = 4 − x v`a f (x, y) = x2 + (4 − x)2 + x(4 − x) − 5x − 4(4 − x) + 10 = x2 − 5x + 10, ta thu du.o. .c h`am mˆo. t biˆe´n sˆo´ g(x) = x2 − 5x + 10 .c tri. di.a phu.o.ng cu’ a g(x) c˜ung ch´ınh l`a cu. .c tri. c´o diˆe`u kiˆe.n cu’ a
v`a cu.
h`am f (x, y). ´Ap du. ng phu.o.ng ph´ap kha’ o s´at h`am sˆo´ mˆo. t biˆe´n sˆo´ dˆo´i
v´o.i g(x) ta t`ım du.o. .c g(x) c´o cu. · (cid:17) = gmin = g(cid:16) .c tiˆe’u di.a phu.o.ng
15
4 5
2 khi d´o h`am f (x, y) cho (cid:16) , d˜a
diˆe’m tiˆe’u c´o kiˆe.n ta. i Nhu.ng
.c
cu. c´o
3
(cid:17)
2 5
2 ) v`a = (y = 4 − x ⇒ y = 4 − diˆe`u
3
2 5
2 · (cid:17) = , fmin = f (cid:16) 5
2 3
2 15
4
2) T`u. c´ac phu.o.ng tr`ınh r`ang buˆo. c ta c´o z = 1 + x
y = x2 + x + 1 v`a thˆe´ v`ao h`am d˜a cho ta du.o. .c h`am mˆo. t biˆe´n sˆo´ u = f (x, y(x), z(x)) = g(x) = 2x2 + 4x + 2. .c tiˆe’u ta. i x = −1 (khi d´o y = 1,
.c tiˆe’u c´o diˆe`u kiˆe.n ta. i diˆe’m Dˆe˜ d`ang thˆa´y r˘a`ng h`am g(x) c´o cu.
z = 0) v`a do d´o h`am f (x, y, z) c´o cu.
(−1, 1, 0) v`a fmin = f (−1, 1, 0) = 0. N 152 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n .c tri. V´ı du. 5. B˘a`ng phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh Lagrange t`ım cu.
c´o diˆe`u kiˆe.n cu’ a h`am u = x + y + z2 v´o.i diˆe`u kiˆe.n (9.16) (z − x = 1
y − xz = 1 (xem v´ı du. 4, ii)). Gia’ i. Ta lˆa. p h`am Lagrange F (x, y, z) = x + y + z2 + λ1(z − x − 1) + λ2(y − zx − 1) v`a x´et hˆe. phu.o.ng tr`ınh = 1 − λ1 − λ2z = 0 = 1 + λ2 = 0 = 2z + λ1 − λ2x = 0 ∂F
∂x
∂F
∂y
∂F
∂z
ϕ1 = z − x − 1 = 0
ϕ2 = y − xz − 1 = 0.
Hˆe. n`ay c´o nghiˆe.m duy nhˆa´t x = −1, y = 1, z = 0, λ1 = 1 v`a
.c tri. cu’ a λ2 = −1 ngh˜ıa l`a M0(−1, 1, 0) l`a diˆe’m duy nhˆa´t c´o thˆe’ c´o cu.
h`am v´o.i c´ac diˆe`u kiˆe.n r`ang buˆo. c ϕ1 v`a ϕ2. T`u. c´ac hˆe. th´u.c z − x = 1 y − xz = 1 ta thˆa´y r˘a`ng (9.16) x´ac di.nh c˘a. p h`am ˆa’n y(x) v`a z(x) (trong tru.`o.ng
.p n`ay y(x) v`a z(x) dˆe˜ d`ang r´ut ra t`u. (9.16)). Gia’ su.’ thˆe´ nghiˆe.m
ho. 153 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n .c ta c´o y(x) v`a z(x) v`ao hˆe. (9.16) v`a b˘a`ng c´ach lˆa´y vi phˆan c´ac dˆo` ng nhˆa´t
th´u.c thu du.o. (dz − dx = 0 (dz = dx ⇒ (9.17) dy − xdz − zdx = 0 dy = (x + z)dx. Bˆay gi`o. t´ınh vi phˆan cˆa´p hai cu’ a h`am Lagrange (9.18) d2F = 2(dz)2 − 2λ2dxdz. Thay gi´a tri. λ2 = −1 v`a (9.17) v`ao (9.18) ta thu du.o. .c da. ng to`an phu.o.ng x´ac di.nh du.o.ng l`a d2F = 4dx2. T`u. d´o suy ra h`am d˜a cho c´o cu. .c tiˆe’u c´o diˆe`u kiˆe.n ta. i diˆe’m M0(−1, 1, 0) v`a fmin = 0. N
V´ı du. 6. T`ım gi´a tri. l´o.n nhˆa´t v`a nho’ nhˆa´t cu’ a h`am f (x, y) = x2 + y2 − xy + x + y trong miˆe`n D = {x 6 0, y 6 0, x + y > −3}. Gia’ i. Miˆe`n D d˜a cho l`a tam gi´ac OAB v´o.i dı’nh ta. i A(−3, 0), B(0, −3) v`a O(0, 0). i) T`ım c´ac diˆe’m d`u.ng: f 0
x = 2x − y + 1 = 0
f 0
y = 2y − x + 1 = 0 T`u. d´o x = −1, y = −1. Vˆa. y diˆe’m d`u.ng l`a M(−1, −1).
Ta. i diˆe’m M ta c´o: f (M) = f (−1, −1) = −1. 154 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n ii) Ta c´o xx(−1, −1) = 2
xy(−1, −1) = −1
yy (−1, −1) = 2. A = f 00
B = f 00
C = f 00 Vˆa.y AC − B2 = 4 − 1 = 3 > 0, nˆen h`am c´o biˆe.t th´u.c AC − B2 > 0
.c tiˆe’u di.a phu.o.ng v`a v`a A = 2 > 0. Do d´o ta. i diˆe’m M n´o c´o cu.
fmin = −1. iii) Kha’ o s´at h`am trˆen biˆen cu’ a miˆe`n D.
+) Khi x = 0 ta c´o f = y2 + y. Dˆo´i v´o.i h`am mˆo. t biˆe´n f = y2 + y, −3 6 y 6 0 ta c´o (fln)(cid:12) = 6 ta. i diˆe’m (0, −3) (cid:12)x=0 (cid:17). = (fnn)(cid:12) ta. i diˆe’m (cid:16)0, − −1
4 1
2 (cid:12)x=0 +) Khi y = 0 ta c´o h`am mˆo. t biˆe´n f = x2 + x, −3 6 x 6 0 v`a .:
tu.o.ng tu. (fln)(cid:12) , 0(cid:17). (fnn)(cid:12) −1
4 (cid:12)y=0 = 6 ta. i diˆe’m (0, −3)
1
ta. i diˆe’m (cid:16) −
(cid:12)y=0 =
2 +) Khi x + y = −3 ⇒ y = −3 − x ta c´o f (x) = 3x2 + 9x + 6 v`a (cid:17) , − (fnn)(cid:12) 3
2 3
2 −3
4 (fln)(cid:12) ta. i diˆe’m (cid:16) −
(cid:12)x+y=−3 =
(cid:12)x+y=−3 = 6 ta. i diˆe’m (0, −3) v`a (−3, 0). .c dˆo´i v´o.i f ta kˆe´t luˆa. n fln = 6 ta. i iv) So s´anh c´ac gi´a tri. thu du.o. (0, −3) v`a (−3, 0) v`a gi´a tri. fnn = −1 ta. i diˆe’m d`u.ng (−1, −1). B `AI T ˆA. P 155 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n .c tri. cu’ a c´ac h`am sau dˆay H˜ay t`ım cu. 1. f = 1 + 6x − x2 − xy − y2. (DS. fmax = 13 ta. i diˆe’m (4, −2)) 2. f = (x − 1)2 + 2y2. (DS. fmin = 0 ta. i diˆe’m (1, 0)) 3. f = x2 + xy + y2 − 2x − y. (DS. fmin = −1 ta. i diˆe’m (1, 0)) 4. f = x3y2(6 − x − y) (x > 0, y > 0). (DS. fmax = 108 ta. i diˆe’m (3, 2)) 5. f = 2x4 + y4 − x2 − 2y2. (DS. fmax = 0 ta. i diˆe’m (0, 0), , 1(cid:17) fmin = − , −1(cid:17) v`a M2(cid:16) ta. i c´ac diˆe’m M1(cid:16) , −1(cid:17)) fmin = − , −1(cid:17) v`a M4(cid:16) ta. i c´ac diˆe’m M3(cid:16) 1
2
1
2 −1
2
−1
2 9
8
9
8 6. f = (5x + 7y − 25)e−(x2+xy+y2 ). (DS. (cid:17)) , fmax = 3−13 ta. i diˆe’m M1(1, 3),
fmin = −26e−1/52 ta. i diˆe’m M2(cid:16) −1
26 −3
26 7. f = xy + + , x > 0, y > 0. (DS. fmin = 30 ta. i diˆe’m (5, 2)) 50
x 20
y 8. f = x2 + xy + y2 − 6x − 9y. √ y − x2 − y + 6x + 3. 9. f = x (DS. fmin = −21 ta. i diˆe’m (1, 4))
(DS. fmax = 15 ta. i diˆe’m (4, 4)) √ 10. f = (x2 + y) ey. (DS. fmin = − ta. i (0, −2)) 2
e 11. f = 2 + (x − 1)4(y + 1)6. (DS. fmin = 2 ta. i diˆe’m (1, −1)) Chı’ dˆa˜n. Ta. i diˆe’m M0(1, −1) ta c´o ∆(M0) = 0. Cˆa` n kha’ o s´at dˆa´u cu’ a f (M) − f (M0) = f (1 + ∆x, −1 + ∆y) − f (1, −1). 12. f = 1 − (x − 2)4/5 − y4/5. (DS. fmax = 1 ta. i diˆe’m (2, 0)) Chı’ dˆa˜n. Ta. i diˆe’m (2, 0) h`am khˆong kha’ vi. Kha’ o s´at dˆa´u cu’ a f (M) − f (M0), M0 = (2, 0). 156 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n T`ım cu. .c tri. c´o diˆe`u kiˆe.n cu’ a c´ac h`am sau dˆay , (DS. fmax = ta. i diˆe’m (cid:16) 13. f = xy v´o.i diˆe`u kiˆe.n x + y = 1.
1
(cid:17))
4 1
2 1
2 14. f = x + 2y v´o.i diˆe`u kiˆe.n x2 + y2 = 5. + = 1. 15. f = x2 + y2 v´o.i diˆe`u kiˆe.n , (cid:17)) (DS. fmin = ta. i diˆe’m (cid:16) (DS. fmax = 5 ta. i diˆe’m (1, 2))
x
2
18
13 36
13 y
3
12
13 16. f = x − 2y + 2z v´o.i diˆe`u kiˆe.n x2 + y2 + z2 = 9. (DS. fmin = −9 ta. i diˆe’m (−1, 2, −2); fmax = 9 ta. i (1, −2, 2).) (cid:17)) , (DS. fmax = ta. i diˆe’m (cid:16) 17. f = xy v´o.i diˆe`u kiˆe.n 2x + 3y = 5.
25
24 5
4 5
6 + = 1. 18. 1) f = x2 + y2 v´o.i diˆe`u kiˆe.n r`ang buˆo. c x
4 y
3 (cid:17)) , ta. i (cid:16) 36
25 48
25 144
25 (DS. fmin =
2) f = exy v´o.i diˆe`u kiˆe.n x + y = 1. (cid:17)) , 1
2 1
2 (DS. fmax = e1/4 ta. i diˆe’m (cid:16)
Chı’ dˆa˜n. C´o thˆe’ su.’ du. ng phu.o.ng ph´ap khu.’ biˆe´n. 19. f = x2 + y2 + 2z2 v´o.i diˆe`u kiˆe.n x − y + z = 1.
(DS. fmin = 0, 4 ta. i diˆe’m (0, 4; −0, 4; 0, 2)) 20. f = x3 + y2 − z3 + 5 v´o.i diˆe`u kiˆe.n x + y − z = 1. , , (cid:17)) ta. i diˆe’m (cid:16)− (DS. fmin = 5 ta. i diˆe’m (0, 0, 0) v`a fmax = 7 10
27 4
3 8
3 4
3 157 9.3. Cu. .c tri. cu’ a h`am nhiˆe`u biˆe´n (cid:17) , , , , , , (cid:17); (cid:16) (cid:17); (cid:16) (DS. fmax = 4 ta. i (cid:16) 4
27 4
3 4
3 7
3 4
3 4
3 7
3 4
3 7
3 21. f = xyz v´o.i c´ac diˆe`u kiˆe.n x + y + z = 5, xy + yz + zx = 8.
4
3
fmin = 4 ta. i (2, 2, 1); (2, 1, 2); (1, 2, 2)) T`ım gi´a tri. l´o.n nhˆa´t v`a nho’ nhˆa´t cu’ a c´ac h`am sˆo´ sau. .c gi´o.i ha. n bo.’ i c´ac doa. n (DS. fln = ta. i diˆe’m (1, 2); fnn = −128 ta. i diˆe’m (4, 2)). 22. f = x2y(2 − x − y), D l`a tam gi´ac du.o.
th˘a’ ng x = 0, y = 0, x + y = 6.
1
4 23. f = x + y, D = {x2 + y2 6 1}. √ √ (cid:17); , (DS. fln = √
2
2 2 ta. i diˆe’m biˆen (cid:16)
√ (cid:17)). , − fnn = − 2 ta. i diˆe’m biˆen (cid:16) − 2
2
√
2
2 √
2
2 24. T`u. mo. i tam gi´ac c´o chu vi b˘a`ng 2p, h˜ay t`ım tam gi´ac c´o diˆe.n t´ıch
l´o.n nhˆa´t. Chı’ dˆa˜n. D˘a. t a = x, b = y ⇒ c = 2p − x − y v`a ´ap du. ng cˆong th´u.c Heron S = pp(p − x)(p − y)(x + y − p) (DS. Tam gi´ac dˆe`u). 25. X´ac di.nh gi´a tri. l´o.n nhˆa´t v`a nho’ nhˆa´t cu’ a h`am f = x2 − y2, D = {x2 + y2 6 1} (DS. fln = 1 ta. i (1, 0) v`a (−1, 0);
fnn = −1 ta. i (0, 1) v`a (0, −1)).
26. X´ac di.nh gi´a tri. l´o.n nhˆa´t v`a nho’ nhˆa´t cu’ a h`am f = x3 − y3 − 3xy, D = {0 6 x 6 2, −1 6 y 6 2}. 158 Chu.o.ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe`u biˆe´n (DS. fln = 13 ta. i diˆe’m (2, −1);
fnn = −1 ta. i diˆe’m (1, 1) v`a (0, −1)).Chu.o.ng 8
Ph´ep t´ınh vi phˆan h`am mˆo. t
biˆe´n
8.1 D- a. o h`am
8.1.1 D- a. o h`am cˆa´p 1
Gia’ su.’ h`am y = f (x) x´ac di.nh trong δ-lˆan cˆa. n cu’ a diˆe’m x0 (U (x0; δ) =
{x ∈ R : |x − x0| < δ) v`a ∆f (x0) = f (x0 + ∆x) − f (x0) l`a sˆo´ gia cu’ a
n´o ta. i diˆe’m x0 tu.o.ng ´u.ng v´o.i sˆo´ gia ∆x = x − x0 cu’ a dˆo´i sˆo´.
8.2 Vi phˆan
8.3 C´ac di.nh l´y co. ba’ n vˆe` h`am kha’ vi.
Quy t˘a´c l’Hospital. Cˆong th´u.c Tay-
lor
Chu.o.ng 9
Ph´ep t´ınh vi phˆan h`am
nhiˆe`u biˆe´n
9.1 D- a. o h`am riˆeng
9.2 Vi phˆan cu’ a h`am nhiˆe`u biˆe´n
9.3 Cu.
.c tri. cu’ a h`am nhiˆe`u biˆe´n
.c tri.
.c tiˆe’u di.a phu.o.ng) b˘a`ng
H`am f (x, y) c´o cu.
f (x0, y0) ta. i diˆe’m M0(x0, y0) ∈ D nˆe´u tˆo` n ta. i δ-lˆan cˆa. n cu’ a diˆe’m M0
sao cho v´o.i mo. i diˆe’m M 6= M0 thuˆo. c lˆan cˆa. n ˆa´y ta c´o
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