SOLUTION —
$f(x)=x(x-1) \sin x-\left(x^3-2 x^2\right) \cos x-x^3 \tan x$
$=x^2 \sin x-x^3 \cos x-x^3 \tan x+2 x^2 \cos x-x \sin x$ $\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^2}$
$=\lim _{x \rightarrow 0}\left(\sin x-x \cos x-x \tan x+2 \cos x-\frac{\sin x}{x}\right)$ $=0-0-0+2-1=1$
So, The correct option will be (D).