Question

If $L=\underset{x\to \frac{\pi }{4}}{\lim }\frac{(1-\tan x)(1-\sin 2x)}{(1+\tan x)(\pi -4x{)}^3}$, then the value of $40L$ is equal to

Answer 1

Answer: 1.25

$L=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\tan \left(\frac{\pi }{4}-x\right)(1-\sin 2x)}{4\left(\frac{\pi }{4}-x\right)(\pi -4x{)}^2}$
$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1}{4}\frac{(1-\sin 2x)}{(\pi -4x{)}^2}$
$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1}{4}\frac{(-\cos 2x)\cdot 2}{2(\pi -4x)(-4)}$
(applying L-Hospital rule)
$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1}{16}\frac{\cos (2x)}{(\pi -4x)}$
$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1}{16}\frac{(-\sin 2x)\times 2}{-4}$
(again applying L-Hospital rule)
$=\frac{1}{16}\left(\frac{-1}{-4}\right)\times 2$
$\Rightarrow L=\frac{1}{32}$