Question

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

Answer 1

The given function is $f\left(x\right)=\frac{1-\tan x}{4x-\mathrm{\pi }},x\ne \frac{\mathrm{\pi }}{4},x\in \left[0,\frac{\mathrm{\pi }}{2}\right].$

It is given that, the function f(x) is continuous in $\left[0,\frac{\mathrm{\pi }}{2}\right]$. So, the function is continuous at $x=\frac{\mathrm{\pi }}{4}$.

$\therefore f\left(\frac{\mathrm{\pi }}{4}\right)$

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

$=\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\tan x}{4x-\mathrm{\pi }}$

Put $x=\frac{\mathrm{\pi }}{4}+h$

When $x\to \frac{\mathrm{\pi }}{4},h\to 0$

So,

$f\left(\frac{\mathrm{\pi }}{4}\right)$
$=\underset{h\to 0}{\lim }\frac{1-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)}{4\left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)-\mathrm{\pi }}$

$=\underset{h\to 0}{\lim }\frac{1-{\displaystyle \frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan h}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan h}}}{4h}$

$=\underset{h\to 0}{\lim }\frac{1-\tan h-1-\tan h}{4h\left(1-\tan h\right)}\left(\tan \frac{\mathrm{\pi }}{4}=1\right)$

$=\underset{h\to 0}{\lim }\frac{-2\tan h}{4h\left(1-\tan h\right)}$

$=-\frac{1}{2}\times \underset{h\to 0}{\lim }\frac{\tan h}{h}\times \frac{1}{\underset{h\to 0}{\lim }\left(1-\tan h\right)}$
$=-\frac{1}{2}\times 1\times \frac{1}{\left(1-0\right)}\left(\underset{x\to 0}{\lim }\frac{\tan x}{x}=1\right)$
$=-\frac{1}{2}$

Thus, the value of $f\left(\frac{\mathrm{\pi }}{4}\right)$ is $-\frac{1}{2}$.


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