#### Answer

$$\frac{1}{\pi }$$

#### Work Step by Step

Given $$\lim_{x\to0 } \frac{\sin x}{\sin \pi x}$$
Since
\begin{align*}
\lim_{x\to0 } \frac{\sin x}{\sin \pi x}&=\lim_{x\to0 } \frac{\sin x}{\sin \pi x}\frac{x}{x}\\
&=\lim_{x\to0 } \frac{\sin x}{ x}\frac{x}{\sin \pi x}\\
&=\lim_{x\to0 } \frac{\sin x}{ x}\lim_{x\to0 } \frac{1}{\frac{\sin \pi x}{x}}\\
&=\lim_{x\to0 } \frac{\sin x}{ x}\left(\frac{\lim_{x\to0 }1}{\pi\lim_{\pi x\to0 }\frac{\sin \pi x}{\pi x}}\right)\\
&= \frac{1}{\pi}
\end{align*}