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: 22

Answer

$\dfrac{1}{2}$

Work Step by Step

Consider: $\lim\limits_{x \to \dfrac{\pi}{2}}f(x)=\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{\ln (\csc)}{(x-(\dfrac{\pi}{2}))^2}$ We need to check that the limit has an indeterminate form. Thus, $f(\dfrac{\pi}{2})=\dfrac{0}{0}$ 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 $\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{-\cot x}{2x-\pi}=\dfrac{0}{0}$ Need to apply L-Hospital's rule again. $\lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{\csc^2 \dfrac{\pi}{2}}{2}=\dfrac{1}{2}$
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.