$\lim_{x\to0} \dfrac{x \tan x}{(1-\cos x)}$

Given limit is indeterminate of type $\frac{0}{0}$.

(DL Rule) Differentiate numerator and denominator:

$\displaystyle \lim_{x\to0}\frac{x\tan x}{1-\cos x}=\lim_{x\to0}\frac{\tan x+x\sec^2 x}{\sin x}$

As $x\to0$, $\tan x\approx x$, $\sin x\approx x$, $\sec^2 x\approx1$

$\Rightarrow \dfrac{x+x}{x}=2$

Answer: $\boxed{2}$ ✅