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: 7



Work Step by Step

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