Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 9



Work Step by Step

Since $-1\leq \sin \frac{\pi}{x-1}\leq 1$, then we have $$-(x-1)\leq (x-1)\sin\frac{\pi}{x-1}\leq (x-1).$$ Moreover, $\lim\limits_{x \to 1}(x-1)=\lim\limits_{x \to 1}-(x-1)=0$. Then by the Squeeze Theorem, we have $$\lim\limits_{x \to 1}(x-1)\sin\frac{\pi}{x-1}=0.$$
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