Answer
$\dfrac{1}{\ln 2}$
Work Step by Step
Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{x 2^x}{2^x-1}$
We need to check that the limit has an indeterminate form.
Thus, $f(0)=\dfrac{(0) 2^{0}}{2^{0}-1}=\dfrac{0}{0}$
The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
Then
$\lim\limits_{x \to 0}\dfrac{2^x+x 2^x \ln 2}{2^x \ln 2}=\dfrac{2^{0}+0}{2^{0} \ln 2}=\dfrac{1}{\ln 2}$