Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - Exercises: 15

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, $$\lim\limits_{x\to\pi^-}\ln(\sin x)=-\infty$$
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