Answer
$$\lim_{x\to\infty}\frac{\sin2x}{x}=0$$
Work Step by Step
We know already that $$-1\le \sin2x\le1$$
So, $$-\frac{1}{x}\le\frac{\sin2x}{x}\le\frac{1}{x}$$
As $x\to\infty$, both $1/x$ and $-1/x$ will approach $0$. So $\lim_{x\to\infty}(-1/x)=\lim_{x\to\infty}(1/x)=0$
Therefore, according to Squeeze Theorem, we conclude that $$\lim_{x\to\infty}\frac{\sin2x}{x}=0$$
