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: 49

Answer

The solution is $$\lim_{x\to\pi/2^-}\left(\frac{\pi}{2}-x\right)\sec x=1.$$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.