## Calculus (3rd Edition)

$\lim\limits_{x \to 0}\frac{sin2x}{x}$= $\lim\limits_{x \to 0}[\frac{sin2x}{2x}.2]$ = 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