Answer
2
Work Step by Step
$\lim\limits_{x \to 0}\frac{sin2x}{x}$= $\lim\limits_{x \to 0}[\frac{sin2x}{2x}.2]$
= 2.$\lim\limits_{2x \to 0}[\frac{sin2x}{2x}]$ (as x tends to 0, 2x also tends to 0)
=2•1 (As $ \lim\limits_{x \to 0}\frac{sinx}{x}=1$)
= 2
