Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 97: 9


$$\lim _{x \rightarrow \infty} \frac{\sin 2 x}{x}=0$$

Work Step by Step

Since $-1 \leq \sin 2 x \leq 1$ for all $ x \in \mathbb{R}$ then $$-\left|\frac{1}{x}\right| \leq \frac{\sin 2 x}{x} \leq\left|\frac{1}{x}\right|$$ for all $ x \in \mathbb{R} \backslash\{0\}$ . Hence, since $\lim _{x \rightarrow \infty}-\left|\frac{1}{x}\right|=0=\lim _{x \rightarrow \infty}\left|\frac{1}{x}\right|$ we have by the sandwich theorem that $$\lim _{x \rightarrow \infty} \frac{\sin 2 x}{x}=0$$
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