Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 46

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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