Evaluate the Given limit:

$\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

It is seen that $x \rightarrow \pi \Rightarrow(\pi-x) \rightarrow 0$

$\therefore \lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}=\frac{1}{\pi} \lim _{(\pi-x) \rightarrow 0} \frac{\sin (\pi-x)}{(\pi-x)}$

$=\frac{1}{\pi} \times 1 \quad\left[\lim _{y \rightarrow 0} \frac{\sin y}{y}=1\right]$

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