Question

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

Answer 1

Answer: 0.5

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