Let $f\left(x\right)=\frac{1-\tan x}{4x-\pi },x\ne \frac{\pi }{4},x\in \left[0,\frac{\pi }{2}\right].$ If $f\left(x\right)$ is continuous in $[0,\pi /2],$ then $f(\pi /4)$ is______

Given that, $f\left(x\right)=\frac{1-\tan x}{4x-\pi },x\ne \frac{\pi }{4},x\in \left[0,\frac{\pi }{2}\right].$

and $f\left(x\right)$ is continuous in $[0,\pi /2],$ 

$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1-\tan x}{4x-\pi }$

 By applying L– H rule get required value

$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{\lim }\frac{-se{c}^{2}x}{4}=-0.5$.