Answer
$\dfrac{1}{\ln 2}$
Work Step by Step
Let $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{x 2^x}{2^x-1}$
But $f(0)=\dfrac{(0) 2^{0}}{2^{0}-1}=\dfrac{0}{0}$
This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$
This implies that $\lim\limits_{x \to 0}\dfrac{(2^x)+)(x) (2^x) (\ln 2)}{(2^x) (\ln 2)}=\dfrac{2^{0}+0}{2^{0} \ln 2}$
Thus, $\dfrac{1+0}{(1) \ln 2}=\dfrac{1}{\ln 2}$
