Answer
$$\lim\limits_{x\to\pi^-}\ln(\sin x)=-\infty$$
Work Step by Step
$$A=\lim\limits_{x\to\pi^-}\ln(\sin x)$$
As $x\to\pi^-$, $\ln(\sin x)$ approaches $-\ln(\sin \pi)=-\ln0=-\infty$
Therefore, $$\lim\limits_{x\to\pi^-}\ln(\sin x)=-\infty$$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 2 - Review - Exercises - Page 168: 15
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