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: 14



Work Step by Step

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