Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 35



Work Step by Step

We have $$ \lim _{x \rightarrow \pi/2}\left(\sec x- \tan x\right)=\lim _{x \rightarrow \pi/2}\left(\frac{1}{\cos x}- \frac{\sin x}{\cos x}\right)\\ =\lim _{x \rightarrow \pi/2}\left(\frac{1-\sin x}{\cos x}\right)=\frac{0}{0} $$ is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \pi/2}\left(\frac{1-\sin x}{\cos x}\right)=\lim _{x \rightarrow \pi/2}\left(\frac{-\cos x}{\sin x}\right)=\frac{0}{1}=0.$$
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