University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 19

Answer

$\dfrac{1}{4}$

Work Step by Step

Consider: $\lim\limits_{\theta \to \dfrac{\pi}{2}}f(x)=\lim\limits_{ \theta \to \dfrac{\pi}{2}}\dfrac{1-\sin \theta }{1+\cos (2\theta)}$ Now, $f(\dfrac{\pi}{2})=\dfrac{1-\sin (\dfrac{\pi}{2}) }{1+\cos (2 (\dfrac{\pi}{2}))}=\dfrac{1-1}{1 +(-1)}=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Then: $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{- \cos \theta}{-2 \sin 2\theta}=\dfrac{- \cos \dfrac{\pi}{2}}{-2 \sin (2 \dfrac{\pi}{2})}=\dfrac{0}{0}$ We need to apply L-Hospital's rule again: $\lim\limits_{\theta \to \dfrac{\pi}{2}}\dfrac{ \sin \theta}{-4 \cos 2\theta}=\dfrac{ \sin \dfrac{\pi}{2}}{-4 \cos 2 (\dfrac{\pi}{2})}=\dfrac{1}{4}$
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.