Step-by-step Solution:

\[ \lim_{x\rightarrow0}\frac{x\tan x}{(1-\cos x)} \] Rewriting \(\tan x = \frac{\sin x}{\cos x}\): \[ = \lim_{x\rightarrow0} \frac{x \cdot \frac{\sin x}{\cos x}}{(1-\cos x)} \] \[ = \lim_{x\rightarrow0} \frac{x\sin x}{\cos x(1-\cos x)} \] Using the identity \(\sin^2 x = 1 - \cos^2 x\): \[ = \lim_{x\rightarrow0} \frac{x(1 - \cos^2 x)}{\sin x \cos x (1 - \cos x)} \] Factorizing \(1 - \cos^2 x\) as \((1 - \cos x)(1 + \cos x)\): \[ = \lim_{x\rightarrow0} \frac{x(1+\cos x)(1-\cos x)}{\sin x \cos x (1-\cos x)} \] Canceling \( (1 - \cos x) \) from numerator and denominator: \[ = \lim_{x\rightarrow0} \frac{x(1+\cos x)}{\sin x \cos x} \] Splitting into two limits: \[ = \lim_{x\rightarrow0} \frac{x}{\sin x} \cdot \lim_{x\rightarrow0} \frac{1+\cos x}{\cos x} \] Since \(\lim_{x\rightarrow0} \frac{x}{\sin x} = 1\) and \(\lim_{x\rightarrow0} \frac{1+\cos x}{\cos x} = 2\), we get: \[ = 1 \times 2 = 2 \] Thus, the correct answer is: \[ \lim_{x\rightarrow0}\frac{x\tan x}{(1-\cos x)} = 2 \]