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: 21

Answer

The solution is $$\lim_{u\to \pi/4}\frac{\tan u-\cot u}{u-\pi/4}=4.$$

Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{u\to \pi/4}\frac{\tan u-\cot u}{u-\pi/4}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{u\to \pi/4}\frac{(\tan u+\cot u)'}{(u-\pi/4)'}=\lim_{u\to \pi/4}\frac{\frac{1}{\cos^2 u}-\left(-\frac{1}{\sin^2 u}\right)}{1}=\lim_{u\to \pi/4}\left(\frac{1}{\cos^2 u}+\frac{1}{\sin^2 u}\right)=\lim_{u\to\pi/4}\frac{\cos^2 u+\sin^2u}{\cos^2 u \sin^2 u}=\lim_{u\to\pi/4}\frac{1}{\cos^2 u\sin^2 u}=\frac{1}{\cos^2\pi/4\sin^2\pi/4}=\frac{1}{(\sqrt{2}/2)^4}=4..$$
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