Question

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

Answer 1

We have,

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

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

So, apply L-hospital rule

${lim}_{x\to 1}\left(\frac{0-\sin \pi x\left(\pi \right)}{2\tan \pi x\left({\sec }^2\pi x\right)\left(\pi \right)}\right)$

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

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

${lim}_{x\to 1}\left(\frac{-{\cos }^3\pi x}{2}\right)$

$=-\frac{{\left(\cos \pi \right)}^3}{2}$

$=-\frac{{(-1)}^3}{2}$

$=\frac{1}{2}$

Hence, this is the answer.