Answer: $\mathbf C$
Solution:
Let $$\mathrm {I = \underset{x\to0}{\lim}\frac{\tan^2x - x\tan x}{\sin x}}$$
On substituting $\mathrm x = 0$, we can see that this integral is of the form, $\color {blue} {\mathbf{\frac{0}{0}}}$ i.e., $\color {blue} {\text {indeterminate form}}$
So, we can apply l'hospital's rule:
$$\therefore \mathrm {I = \underset{x\to0}{\lim}\frac{\frac{d}{dx}(\tan^2x-x\tan x)}{\frac{d}{dx}\sin x} }$$
$$\mathrm {\;\;\left(\because \frac{d}{dx}\tan x = \sec^2x,\; \frac{d}{dx}\cos x = \sin x\right)}$$
$$\mathrm {\Rightarrow \mathrm I = \underset{x \to 0}{\lim}\frac{\left (2\tan x \sec^2 x-(x.\sec^2x + 1.\tan x) \right)}{\cos x}}$$
$$\mathrm {\Rightarrow \mathrm I = \underset {x \to 0}{\lim}\left \{\left (2\frac{\sin x}{\cos x}.\frac{1}{\cos^2x} - \left(\frac{x}{\cos^2x}+\frac{\sin x}{\cos x}\right)\right)\right \}.\frac{1}{\cos x}}$$
Now, substitute $\mathrm x = 0$, we get:
$$\left (\text{Also}\;\because \cos 0 = 1, \;\sin 0 = 0\right)$$
$$\therefore \mathrm I = 0$$
Answer: $\therefore \mathbf C$ is the correct option.