Question

Evaluate the limit:
$\underset{x\to 1}{lim}\frac{1-\frac{1}{x}}{\sin \pi (x-1)}$

Answer 1

We have:
$\underset{x\to 1}{lim}\frac{1-\frac{1}{x}}{\sin \pi (x-1)}$

$=\underset{x\to 1}{lim}\frac{x-1}{x\sin \pi (x-1)}$

Put $x=1+h$,
If $x\to 1$, then $h\to 0$

$=\underset{h\to 1}{lim}\frac{1+h-1}{(1+h)\sin \pi (1+h-1)}$

$=\underset{h\to 1}{lim}\frac{h}{(1+h)\sin \pi h}$

$=\underset{h\to 1}{lim}\frac{1}{(1+h)\times \pi \times \frac{\sin \pi h}{\pi h}}$

$=\frac{1}{(1+0)\times \pi \times 1}$
$[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]=\frac{1}{\pi }$

Therefore,
$\underset{x\to 1}{lim}\frac{1-\frac{1}{x}}{\sin \pi (x-1)}=\frac{1}{\pi }$