L'Hopital's Rule for Indeterminate Limits, Schemes and Mind Maps of Pre-Calculus

Download L'Hopital's Rule for Indeterminate Limits and more Schemes and Mind Maps Pre-Calculus in PDF only on Docsity! 31 L’Hopital’s Rule 31.1 Limit of indeterminate type Some limits for which the substitution rule does not apply can be found by using inspection. For instance, lim x→0 cosx x2 ( about 1 small pos. ) =∞ On the other hand, we have seen (8) that inspection cannot be used to find the limit of a fraction when both numerator and denominator go to 0. The examples given were lim x→0+ x2 x , lim x→0+ x x2 , lim x→0+ x x . In each case, both numerator and denominator go to 0. If we had a way to use inspection to decide the limit in this case, then it would have to give the same answer in all three cases. Yet, the first limit is 0, the second is ∞ and the third is 1 (as can be seen by canceling x’s). We say that each of the above limits is indeterminate of type 0 0 . A useful way to remember that one cannot use inspection in this case is to imagine that the numerator going to 0 is trying to make the fraction small, while the denominator going to 0 is trying to make the fraction large. There is a struggle going on. In the first case above, the numerator wins (limit is 0); in the second case, the denominator wins (limit is ∞); in the third case, there is a compromise (limit is 1). Changing the limits above so that x goes to infinity instead gives a different indeterminate type. In each of the limits lim x→∞ x2 x , lim x→∞ x x2 , lim x→∞ x x . both numerator and denominator go to infinity. The numerator going to infinity is trying to make the fraction large, while the denominator going to infinity is trying to make the fraction small. Again, there is a struggle. Once again, we can cancel x’s to see that the first limit is∞ (numerator wins), the second limit is 0 (denominator wins), and the third limit is 1 (compromise). The different answers show that one cannot use inspection in this case. Each of these limits is indeterminate of type ∞∞ . Sometimes limits of indeterminate types 0 0 or ∞∞ can be determined by using some algebraic technique, like canceling between numerator and denominator 1 31 L’HOPITAL’S RULE 2 as we did above (see also 12). Usually, though, no such algebraic technique suggests itself, as is the case for the limit lim x→0 x2 sinx , which is indeterminate of type 0 0 . Fortunately, there is a general rule that can be applied, namely, l’Hôpital’s rule. 31.2 L’Hôpital’s rule L’Hôpital’s rule. If the limit lim f(x) g(x) is of indeterminate type 0 0 or ±∞±∞ , then lim f(x) g(x) = lim f ′(x) g′(x) , provided this last limit exists. Here, lim stands for lim x→a , lim x→a± , or lim x→±∞ . The pronunciation is lō-pē-täl. Evidently, this result is actually due to the mathematician Bernoulli rather than to l’Hôpital. The verification of l’Hôpital’s rule (omitted) depends on the mean value theorem. 31.2.1 Example Find lim x→0 x2 sinx . Solution As observed above, this limit is of indeterminate type 0 0 , so l’Hôpital’s rule applies. We have lim x→0 x2 sinx ( 0 0 ) l’H = lim x→0 2x cosx = 0 1 = 0, where we have first used l’Hôpital’s rule and then the substitution rule. The solution of the previous example shows the notation we use to indicate the type of an indeterminate limit and the subsequent use of l’Hôpital’s rule. 31.2.2 Example Find lim x→−∞ 3x− 2 ex2 . 31 L’HOPITAL’S RULE 5 The strategy for handling this type is to rewrite the product as a quotient and then use l’Hôpital’s rule. 31.5.1 Example Find lim x→∞ xe−x. Solution As noted above, this limit is indeterminate of type ∞· 0. We rewrite the expression as a fraction and then use l’Hôpital’s rule: lim x→∞ xe−x = lim x→∞ x ex (∞ ∞ ) l’H = lim x→∞ 1 ex ( 1 large pos. ) = 0. 31.5.2 Example Find lim x→0+ (cot 2x)(sin 6x). Solution First lim x→0+ cot 2x = lim x→0+ cos 2x sin 2x ( about 1 small pos. ) =∞. Therefore, the given limit is indeterminate of type∞·0. We rewrite as a fraction and then use l’Hôpital’s rule: lim x→0+ (cot 2x)(sin 6x) = lim x→0+ sin 6x tan 2x ( 0 0 ) l’H = lim x→0+ 6 cos 6x 2 sec2 2x = 6 2 = 3. 31.6 Indeterminate difference Type ∞−∞. The substitution rule cannot be used on the limit lim x→π 2 − (tanx− secx) (∞−∞) since tanπ/2 is undefined. Nor can one determine this limit by using inspec- tion. The first term going to infinity is trying to make the expression large and positive, while the second term going to negative infinity is trying to make the 31 L’HOPITAL’S RULE 6 expression large and negative. There is a struggle going on. We say that this limit is indeterminate of type ∞−∞. The strategy for handling this type is to combine the terms into a single fraction and then use l’Hôpital’s rule. 31.6.1 Example Find lim x→π 2 − (tanx− secx). Solution As observed above, this limit is indeterminate of type ∞−∞. We combine the terms and then use l’Hôpital’s rule: lim x→π 2 − (tanx− secx) = lim x→π 2 − ( sinx cosx − 1 cosx ) = lim x→π 2 − sinx− 1 cosx ( 0 0 ) l’H = lim x→π 2 − cosx − sinx = 0 −1 = 0. 31.6.2 Example Find lim x→1+ ( x x− 1 − 1 lnx ) . Solution The limit is indeterminate of type ∞ −∞. We combine the terms and then use l’Hôpital’s rule: lim x→1+ ( x x− 1 − 1 lnx ) = lim x→1+ x lnx− x+ 1 (x− 1) lnx ( 0 0 ) l’H = lim x→1+ lnx+ x(1/x)− 1 lnx+ (x− 1)(1/x) = lim x→1+ lnx lnx+ 1− 1/x ( 0 0 ) l’H = lim x→1+ 1/x 1/x+ 1/x2 = 1 2 . 31.7 Indeterminate powers Type ∞0. The limit lim x→∞ x1/x (∞0) 31 L’HOPITAL’S RULE 7 cannot be determined by using inspection. The base going to infinity is trying to make the expression large, while the exponent going to 0 is trying to make the expression equal to 1. There is a struggle going on. We say that this limit is indeterminate of type ∞0. The strategy for handling this type (as well as the types 1∞ and 00 yet to be introduced) is to first find the limit of the natural logarithm of the expression (ultimately using l’Hôpital’s rule) and then use an inverse property of logarithms to get the original limit. 31.7.1 Example Find lim x→∞ x1/x. Solution As was noted above, this limit is of indeterminate type ∞0. We first find the limit of the natural logarithm of the given expression: lim x→∞ lnx1/x = lim x→∞ (1/x) lnx (0 · ∞) = lim x→∞ lnx x (∞ ∞ ) l’H = lim x→∞ 1/x 1 = 0. Therefore, lim x→∞ x1/x = lim x→∞ eln x 1/x = e0 = 1, where we have used the inverse property of logarithms y = eln y and then the previous computation. (The next to the last equality also uses continuity of the exponential function.) Type 1∞. The limit lim x→∞ ( 1 + 1 x )x (1∞) cannot be determined by using inspection. The base going to 1 is trying to make the expression equal to 1, while the exponent going to infinity is trying to make the expression go to ∞ (raising a number greater than 1 to ever higher powers produces ever larger results). We say that this limit is indeterminate of type 1∞. 31.7.2 Example Find lim x→∞ ( 1 + 1 x )x . Solution As was noted above, this limit is indeterminate of type 1∞. We first