Question

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

Answer 1

We have,

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

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

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

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

$=\underset{h\to 0}{\text{lim}}\frac{1-\cos \pi h}{h^2}$ $[\because \cos (\pi +\theta )=-\cos \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{2{\sin }^2\frac{\pi h}{2}}{h^2}$ $[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=2\underset{h\to 0}{\text{lim}}\frac{{\sin }^2\frac{\pi h}{2}}{4\times \frac{1}{{\pi }^2}\times \frac{{\pi }^2{h}^2}{4}}$

$=\frac{{\pi }^2}{2}\underset{h\to 0}{\text{lim}}\frac{{\sin }^2\frac{\pi h}{2}}{\frac{{\pi }^2{h}^2}{4}}$

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

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


$=\frac{{\pi }^2}{2}$

Therefore,

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