Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{\sin x}} \cr
& {\text{Evaluate the given limit without using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{\sin x}} = \frac{{{e^0} - 1}}{{\sin 0}} = \frac{0}{0} \cr
& {\text{Evaluate using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{\sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{e^x} - 1} \right]}}{{\frac{d}{{dx}}\left[ {\sin x} \right]}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{\cos x}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{\cos x}} = \frac{{{e^0}}}{{\cos 0}} = \frac{1}{1} = 1 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{\sin x}} = 1 \cr} $$
