University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 80

Answer

$2 \pi ^2$

Work Step by Step

Consider $f(x)=\lim\limits_{x \to 4}\dfrac{\sin ^2 (\pi x)}{e^{x-4}+3-x}=\lim\limits_{x \to 4} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$ Thus, $f(0)=\dfrac{0}{0}$ This shows an Inderminate form of the limit, so apply L'Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ Thus, $\lim\limits_{x \to 4}\dfrac{(2 \pi) \sin (\pi x)(\cos \pi x)}{e^{x-4}-1}=\dfrac{0}{0}$ Again apply L'Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ $\lim\limits_{x \to 4}\dfrac{(2 \pi^2) (\cos \pi)}{e^{x-4}}=\dfrac{(2 \pi^2) (\cos 4\pi)}{e^{4-4}}=2 \pi ^2$
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.