$\large \lim_{x->0}\frac{\tan ^{2}x-x\tan x}{sinx}$
dividing numerator and denomerator by x we get,
=$\large \lim_{x->0}\frac{\frac{\tan ^{2}x}{x}-\tan x}{\frac{\sin x}{x}}$
=$\large \frac{\lim_{x->0}\frac{\tan ^{2}x}{x}-\lim_{x->0}\tan x}{\lim_{x->0}\frac{\sin x}{x}}$
now, $\large \lim_{x->0}\frac{\sin x}{x}=1$,$\large \lim_{x->0}\tan x=0$
so we just need to evaluate ,
$\large \lim_{x->0}\frac{\tan ^{2}x}{x}$
=$\large \lim_{x->0}x.\frac{\tan ^{2}x}{x^{2}}$
=$\large \lim_{x->0}x.\lim_{x->0}\frac{\sin ^{2}x}{x^{2}}.\lim_{x->0}\frac{1}{\cos ^{2}x}$
=$0*1*1$
=$0$
so our required answer ,
=$\large \frac{0-0}{1}=0$
correct answer is option C.