Question

Evaluate the given limit :
$\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}$

Answer 1

Given : $\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}$
Substituting $x=\pi$ in the given limit,
$\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}=\underset{x\to \pi }{lim}\frac{\sin (\pi -\pi )}{\pi (\pi -\pi )}$
$=\frac{\sin \left(0\right)}{\pi \left(0\right)}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form.

We need to simplify it,
$\Rightarrow \underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}$
Let $z=\pi -x$
So, when $x\to \pi$
$z\to \pi -\pi$
$z\to 0$
So, our equation becomes
$\underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}=\underset{z\to 0}{lim}\frac{\sin z}{\pi z}$
$=\frac{1}{\pi }\times \underset{z\to 0}{lim}\frac{\sin z}{z}$

$\{\text{Using}\underset{a\to 0}{lim}\frac{\sin a}{a}=1\}$
$=\frac{1}{\pi }\times 1$
$=\frac{1}{\pi }$
$\therefore \underset{x\to \pi }{lim}\frac{\sin (\pi -x)}{\pi (\pi -x)}=\frac{1}{\pi }$