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: 8


See the proof below.

Work Step by Step

Since $-1\leq \sin \frac{1}{x^2}\leq 1$, then we have $$-x\leq x\sin\frac{1}{x^2}\leq x.$$ Moreover, $\lim\limits_{x \to 0}x=\lim\limits_{x \to 0}-x=0$, then by the Squeeze Theorem, we have $$\lim\limits_{x \to 0}x\sin \frac{1}{x^2}=0.$$
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