Chapter 5: Inverse functinons – Section 5.8: Indeterminate forms and l’hospital’s rule

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Nội dung Text: Chapter 5: Inverse functinons – Section 5.8: Indeterminate forms and l’hospital’s rule

SECTION 5.8 INDETERMINATE FORMS AND L’HOSPITAL’S RULE
INDETERMINATE FORMS Suppose we are trying to analyze the behavior of the function ln x F ( x) = x −1 Although F is not defined when x = 1, we need to know how F behaves near 1. In particular, we would like to know the value of the limit ln x lim x 1 x −1 5.8 P2
INDETERMINATE FORMS In computing this limit, we can’t apply Law 5 of limits (Section 1.4) because the limit of the denominator is 0.  In fact, although the limit in Expression 1 exists, its value is not obvious because both numerator and 0 denominator approach 0 and is not defined. 0 5.8 P3
INDETERMINATE FORM —TYPE 0/0 In general, if we have a limit of the form f ( x) lim x a g ( x) where both f(x) → 0 and g(x) → 0 as x → a, then this limit may or may not exist. 0 It is called an indeterminate form of type . 0  We met some limits of this type in Chapter 1. 5.8 P4
INDETERMINATE FORMS For rational functions, we can cancel common factors: x −x 2 x( x − 1) lim 2 = lim x 1 x −1 x 1 ( x + 1)( x − 1) x 1 = lim = x 1 x +1 2 We used a geometric argument to show that: sin x lim =1 x 0 x 5.8 P5
INDETERMINATE FORMS However, these methods do not work for limits such as Expression 1.  Hence, in this section, we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms. Another situation in which a limit is not obvious occurs when we look for a horizontal asymptote of F and need to evaluate the limit ln x lim x x −1 5.8 P6
INDETERMINATE FORMS It isn’t obvious how to evaluate this limit because both numerator and denominator become large as x → ∞. There is a struggle between the two.  If the numerator wins, the limit will be ∞.  If the denominator wins, the answer will be 0.  Alternatively, there may be some compromise—the answer may be some finite positive number. 5.8 P7
INDETERMINATE FORM —TYPE ∞/∞ In general, if we have a limit of the form f ( x) lim x a g ( x) where both f(x) → ∞ (or – ∞) and g(x) → ∞ (or – ∞), then the limit may or may not exist. It is called an indeterminate form of type ∞/∞. 5.8 P8
INDETERMINATE FORMS We saw in Section 1.6 that this type of limit can be evaluated for certain functions—including rational functions—by dividing the numerator and denominator by the highest power of x that occurs in the denominator.  For instance, 1 1− 2 x −1 2 x 1− 0 1 lim 2 = lim = = x 2x + 1 x 1 2+0 2 2+ 2 x 5.8 P9
INDETERMINATE FORMS This method, though, does not work for limits such as Expression 2.  However, L’Hospital’s Rule also applies to this type of indeterminate form. 5.8 P10
L’HOSPITAL’S RULE Suppose f and g are differentiable and g’(x) ≠ 0 on an open interval I that contains a (except possibly at a). Suppose lim f ( x) = 0 and lim g ( x) = 0 x a x a or that lim f ( x) = and lim g ( x) = x a x a  In other words, we have an indeterminate form of 0 type or ∞/∞. 0 Then, f ( x) f '( x) lim = lim x a g ( x) x a g '( x ) if the limit on the right exists (or is ∞ or – ∞). 5.8 P11
NOTE 1 L’Hospital’s Rule says that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives—provided that the given conditions are satisfied.  It is especially important to verify the conditions regarding the limits of f and g before using the rule. 5.8 P12
NOTE 2 The rule is also valid for onesided limits and for limits at infinity or negative infinity.  That is, ‘‘x → a’’ can be replaced by any of the symbols x → a+, x → a–, x →∞, or x → – ∞. 5.8 P13
NOTE 3 For the special case in which f(a) = g(a) = 0, f ’ and g’ are continuous, and g’(a) ≠ 0, it is easy to see why the rule is true. 5.8 P14
NOTE 3 In fact, using the alternative form of the definition of a derivative, we have: f ( x) − f (a) f ( x) − f (a) lim f '( x) f '(a ) x a x−a x−a lim = = = lim x a g '( x ) g '(a ) lim g ( x) − g (a ) x a g ( x) − g (a ) x a x−a x−a f ( x) − f (a) f ( x) = lim = lim x a g ( x) − g (a ) x a g ( x) The general version of l’Hospital’s Rule is more difficult; its proof can be found inAppendix B. 5.8 P15
Example 1 Find lim ln x x 1 x −1 SOLUTION  lim ln x = ln1 = 0 and lim( x − 1) = 0 x 1 x 1  Thus, we can apply l’Hospital’s Rule: d (ln x) ln x dx 1/ x 1 lim = lim = lim = lim = 1 x 1 x −1 x 1 d x 1 1 x 1 x ( x − 1) dx 5.8 P16
Example 2 ex Calculate lim 2 x x SOLUTION We have lim e = lim x = x 2  and x x  So, l’Hospital’s Rule gives: d x (e ) ex e x lim 2 = lim dx = lim x x x d 2 x 2x (x ) dx 5.8 P17
Example 2 SOLUTION As ex → ∞ and 2x → ∞ as x → ∞, the limit on the right side is also indeterminate. However, a second application of l’Hospital’s Rule gives: ex ex ex lim 2 = lim = lim = x x x 2x x 2 5.8 P18
Example 3 ln x Calculate lim 3 x x SOLUTION  As ln x → ∞ and 3 as x x → ∞, l’Hospital’s Rule applies: ln x 1/ x lim 3 = lim 1 −2 / 3 x x x 3x  Notice that the limit on the right side is now 0 indeterminate of type . 0 5.8 P19
Example 3 SOLUTION  However, instead of applying the rule a second time as we did in Example 2, we simplify the expression and see that a second application is unnecessary: ln x 1/ x 3 lim 3 = lim 1 −2 / 3 = lim 3 = 0 x x x 3x x x 5.8 P20