Question

Evaluate the limit:
$\underset{x\to 1}{\text{lim}}\frac{(1-x^2)}{\sin \pi x}$

Answer 1

We have,

$\underset{x\to 1}{\text{lim}}\frac{(1-x^2)}{\sin \pi x}$

Put $x=1+h$,

If $x\to 1$ then $h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{(1-(1+h{)}^2)}{\sin \pi (1+h)}$

$=\underset{h\to 0}{\text{lim}}\frac{(1-(1+h^2+2h\left)\right)}{\sin (\pi +\pi h)}$

$[\because (a+b{)}^2=a^2+b^2+2ab]$

$=\underset{h\to 0}{\text{lim}}\frac{-(h^2+2h)}{-\sin \pi h}$

$[\because \sin (\pi +\theta )=-\sin \theta ]$

Dividing Numerator and Denominator by h

$=\underset{h\to 0}{\text{lim}}\frac{\frac{(h^2+2h)}{h}}{\pi \times \frac{\sin \pi h}{\pi h}}$

$=\frac{1}{\pi }\underset{h\to 0}{\text{lim}}\frac{h+2}{\frac{\sin \pi h}{\pi h}}$ $[\because \underset{h\to 0}{\text{lim}}\frac{\sin h}{h}=1]$

$=\frac{1}{\pi }\times \frac{2}{1}=\frac{2}{\pi }$