Question

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

Answer 1

We have,

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

This is $\frac{0}{0}$ form. So, by using L-Hospital rule,

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

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

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

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

Now, put the limits, we have

$=\frac{-{\cos }^3\pi }{2}$

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

$=\frac{1}{2}$

Hence, the value is $\frac{1}{2}$.