Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 57

Answer

The solution is $$\lim_{\theta\to0}(\tan\theta)^{\cos\theta}=1.$$

Work Step by Step

To solve this limit we will first transform the expression under the limit. Because the natural exponential function and the natural logarithmic function are inverse to each other it holds that $a=e^{\ln a}.$ This gives $(\tan \theta)^{\cos \theta}=e^{\ln (\tan \theta)^{\cos \theta}}=e^{\cos\theta\ln\tan\theta},$ where we also used the logarithmic rule $\ln b^a=a\ln b.$ Because the exponential is continuous it can exchange places with the limit so we get $$L=\lim_{x\to\pi/2^-}(\tan\theta)^{\cos\theta}=\lim_{x\to\pi/2^-}e^{\cos\theta\ln\tan\theta}=e^{\lim_{x\to\pi/2^-}\cos\theta\ln\tan\theta}=e^l,$$ where we denoted $l=\lim_{x\to\pi/2^-}\cos\theta\ln\tan\theta$. Let us calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{x\to\pi/2^-}\cos\theta\ln\tan\theta=\lim_{x\to\pi/2^-}\frac{\ln\tan\theta}{\frac{1}{\cos\theta}}=\left[\frac{\ln\tan\pi/2^-}{\frac{1}{\cos^2\pi/2^-}}\right]=\left[\frac{\ln\infty}{\frac{1}{0^+}}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\pi/2^-}\frac{(\ln\tan\theta)'}{\left(\frac{1}{\cos\theta}\right)'}=\lim_{x\to\pi/2^-}\frac{\frac{1}{\tan\theta}(\tan\theta)'}{-\frac{1}{\cos^2\theta}(\cos\theta)'}=\lim_{x\to\pi/2^-}\frac{\frac{\cos\theta}{\sin\theta}\cdot\frac{1}{\cos^2\theta}}{\frac{\sin\theta}{\cos^2\theta}}=\lim_{x\to\pi/2^-}\frac{\cos\theta}{\sin\theta}=\lim_{x\to\pi/2^-}\cot\theta=\cot(\pi/2)=0.$$ Putting this into the intial limit we get $$L=e^l=e^0=1.$$
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