Question

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

Answer 1

$f(x) = \frac{ 1 -tan\,x}{4x - \pi}$, is continuous in $[0, \frac{\pi}{2}]$
$\therefore f(\frac{\pi}{4}) = \lim\limits_{x \to \frac{\pi}{4}} f(x) = \lim\limits_{x \to \frac{\pi^+}{4}} f(x)$
$= \lim\limits_{h \to 0} f(\frac{\pi}{4} + h)$
$= \lim\limits_{x \to 0} \frac{1 - tan(\frac{\pi}{4} + h)}{4(\frac{\pi}{4} + h) - \pi}, h > 0$
$=\lim\limits_{x \to 0} \frac{1 - \frac{1 + tan\,h}{ 1- tan\,h}}{4h}$
$= \lim\limits_{x \to 0} \frac{-2}{-tanh} . \frac{tan\,h}{4h}$
$ = \frac{-2}{4} = -\frac{1}{2}$
$\therefore | f(\frac{\pi}{4})| = | -\frac{1}{2}| = \frac{1}{2}$