Question

 предел x->0 (cos2x-1)/sinx

Answer 1

$limlimits_{x to 0} frac{cos2x-1}{sinx}\\\ frac{cos2x-1}{sinx} = -frac{1-cos2x}{sinx} =\\ -frac{2(1-cos2x)}{2sinx} = -2frac{sin^2x}{sinx} = -2sinx\\\ limlimits_{x to 0}-2sinx = 0$