Question

If

$f\left(x\right)=\frac{\tan x}{x-\pi }$, then

${lim}_{x\to \pi }f\left(x\right)=?$

Answer 1

We have,

$f\left(x\right)=\frac{\tan x}{x-\pi }$

Since, ${lim}_{x\to \pi }f\left(x\right)$

Now,

${lim}_{x\to \pi }\left(\frac{\tan x}{x-\pi }\right)$

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

So, apply L-Hospital rule,

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

${lim}_{x\to \pi }\left({\sec }^{2}x\right)$

${lim}_{x\to \pi }{\left(\sec x\right)}^2$

$={\left(\sec \pi \right)}^2$

$={(-1)}^2$

$=1$

Hence, this is the answer.