Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 40

Answer

$$\frac{1}{\pi }$$

Work Step by Step

Given $$\lim_{x\to0 } \frac{\sin x}{\sin \pi x}$$ Since \begin{align*} \lim_{x\to0 } \frac{\sin x}{\sin \pi x}&=\lim_{x\to0 } \frac{\sin x}{\sin \pi x}\frac{x}{x}\\ &=\lim_{x\to0 } \frac{\sin x}{ x}\frac{x}{\sin \pi x}\\ &=\lim_{x\to0 } \frac{\sin x}{ x}\lim_{x\to0 } \frac{1}{\frac{\sin \pi x}{x}}\\ &=\lim_{x\to0 } \frac{\sin x}{ x}\left(\frac{\lim_{x\to0 }1}{\pi\lim_{\pi x\to0 }\frac{\sin \pi x}{\pi x}}\right)\\ &= \frac{1}{\pi} \end{align*}
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