Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{2^{\sin x}} - 1}}{{{e^x} - 1}} \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{2^{\sin x}} - 1}}{{{e^x} - 1}} = \frac{{{2^{\sin \left( 0 \right)}} - 1}}{{{e^0} - 1}} \cr
& = \frac{{1 - 1}}{{1 - 1}} \cr
& = \frac{0}{0} \cr
& {\text{the limit is }}\frac{0}{0}{\text{}}{\text{, so we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {{2^{\sin x}} - 1} \right)}}{{d/dx\left( {{e^x} - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{2^{\sin x}}\left( {\ln 2} \right)\left( {\cos x} \right)}}{{{e^x}}} \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{{{2^{\sin \left( 0 \right)}}\left( {\ln 2} \right)\left( {\cos 0} \right)}}{{{e^0}}} \cr
& = \frac{{\ln 2}}{1} \cr
& {\text{then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{2^{\sin x}} - 1}}{{{e^x} - 1}} = \ln 2 \cr} $$
