Question

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

Answer 1

We have,

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

Put $x=1-h,h\to 0$

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

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

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

$=\underset{h\to 0}{\text{lim}}\frac{-h(h-2)}{-\sin \left(2\pi h\right)}$ $[\because \sin (2\pi -\theta )=-\sin \theta ]$

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

$=\frac{(0-2)}{1\times 2\pi }$ $[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1]$

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

Therefore,

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