Answer
$0$
Work Step by Step
$-\dfrac {1}{x}\leq \dfrac {\sin 2x}{x}\leq \dfrac {1}{x}$
$\lim _{x\rightarrow \infty }-\dfrac {1}{x}=0$
$\lim _{x\rightarrow \infty }\dfrac {1}{x}=0$
$\Rightarrow 0\leq \lim _{x\rightarrow \infty }\dfrac {\sin 2x}{x}\leq 0\Rightarrow \lim _{x\rightarrow \infty }\dfrac {\sin 2x}{x}=0$
