Question

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

Answer 1

We have,

${lim}_{x\to 1}\left(\frac{1+\cos \pi x}{\pi (1-x^2)}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 1}\left(\frac{0+(-\sin \pi x)\times \pi }{\pi (0-2x)}\right)$

${lim}_{x\to 1}\left(\frac{\sin \pi x\times \pi }{2x\pi }\right)$

$\Rightarrow \frac{\sin \pi }{2}$

$\Rightarrow 0$

Hence, this is the answer.