Question

${lim}_{x\to \pi /4}\frac{{\sec }^{2}x-2\tan x}{1+\cos 4x}$

Answer 1

$\underset{x\to \frac{\pi }{4}}{lim}\frac{{\sec }^{2}x+2\tan x}{1+\cos 4a}$

$u\mathrm{sing}L-hospitalrule$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{2\mathrm{secsecxtanx}-2se{c}^{2}x}{-4\sin 4x}$

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{2{\sec }^{2}x\left(\tan x-1\right)}{-4\sin 4x}$

Using L' Hospital rule.

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-1}{2}\left[\frac{2\sec x\sec x\tan x\left(\tan x-1+{\sec }^{2}x\times {\sec }^{2}x\right)}{-4\cos 4x}\right]$

$=\frac{-1}{2}\left[\frac{2\times 2\times 1\left(1-1\right)+2\times 2}{-4\times -1}\right]$

$=\frac{-1}{2}\times \frac{4}{4}$
$=\frac{-1}{2}$