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: 53

Answer

The solution is $$\lim_{\theta\to\pi/2^-}(\tan\theta-\sec\theta)=0.$$

Work Step by Step

To solve this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{\theta\to\pi/2^-}(\tan\theta-\sec\theta)=\lim_{\theta\to\pi/2^-}\left(\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos \theta}\right)=\lim_{\theta\to\pi/2^-}\frac{\sin\theta-1}{\cos\theta}=\left[\frac{\sin\frac{\pi}{2}^--1}{\cos\pi/2^-}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{\theta\to\pi/2^-}\frac{(\sin\theta-1)'}{(\cos\theta)'}=\lim_{\theta\to\pi/2^-}\frac{\cos\theta}{-\sin\theta}=\lim_{\theta\to\pi/2^-}-\cot x=\left[-\cot\frac{\pi}{2}^-\right]=0.$$
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