## Thomas' Calculus 13th Edition

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