$ \lim_{x\rightarrow 0} \frac{x 2^x - x}{1-\cos x}$
It forms $\frac{0}{0}$, Apply L' Hospital's Rule
$ \lim_{x\rightarrow 0} \frac{x 2^x\log 2 + 2^x- 1}{0+\sin x}$
Again $\frac{0}{0}$, Apply L'Hospital's rule one more time
$ \lim_{x\rightarrow 0} \frac{(x 2^x\log 2 +2^x) \log 2 + 2^x\log 2- 0}{\cos x}$
$\\ = 2\log 2 = \log 2^2= \log 4$