Answer
$$\frac{m}{n}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin mx}}{{\sin nx}} \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{{\sin m\left( 0 \right)}}{{\sin n\left( 0 \right)}} = \frac{0}{0} \cr
& {\text{Using the lHopital's rule}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\sin mx} \right)}}{{\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\sin nx} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{m\cos mx}}{{n\cos mx}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{m\cos m\left( 0 \right)}}{{n\cos m\left( 0 \right)}} \cr
& = \frac{{m\left( 1 \right)}}{{n\left( 1 \right)}} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin mx}}{{\sin nx}} = \frac{m}{n} \cr} $$
