Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 199: 56

Answer

$$\lim\limits_{x\to0}\csc x\sin(\sin x)=1$$

Work Step by Step

$$A=\lim\limits_{x\to0}\csc x\sin(\sin x)$$ $$A=\lim\limits_{x\to0}\frac{\sin(\sin x)}{\sin x}$$ Let $\sin x=\theta$, then as $x\to0$, $\sin x\to0$, so $\theta\to0$ as well. Therefore, $$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1$$
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