$\lim\limits_{x \rightarrow 0} \frac{1-\cos x}{x\left(2^x-1\right)}$ is equal to

$\frac{1}{2} \log _2 e$

$\lim\limits_{x \rightarrow 0} \frac{1-\cos x}{x\left(2^x-1\right)} . \frac{x}{e^x-1}$

$=\lim\limits_{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{4\left(\frac{x}{2}\right)^2} . \frac{x}{x \log _e 2-1}=\lim\limits_{x \rightarrow 0} \frac{1}{2}\left(\frac{\sin _{\frac{x}{2}}}{\frac{x}{2}}\right)^2 \times \frac{x}{\left(1+x \log _e 2+....\right)-1}=\frac{1}{2} \log _2 e$

Hence (1) is the correct answer.