Answer
$1$
Work Step by Step
$\lim_{x\to0}\frac{x\cdot sin(x)}{2-2cos(x)}=\lim_{x\to0}\frac{2x\cdot sin(x/2)cos(x/2)}{2(1-cos(x))}=\lim_{x\to0}\frac{x\cdot sin(x/2)cos(x/2)}{2sin^2(x/2)}=\lim_{x\to0}\frac{x/2}{sin(x/2)}\frac{cos(x/2)}{1}=(1)(1)=1$