Question

lim x → π/4 [(√2) -cosx-sinx/(4x-π)power 2]​

Answer 1

Answer:

Please see the attachment

Answer 2

Given :

$\lim_{x\to \frac{\pi}{4}}\Big[\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}\Big]$      

To find the solution, let,

$L=\lim_{x\to \frac{\pi}{4}}\Big[\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}\Big]$      $(\frac{0}{0})$ form.

Therefore applying L-Hospital rule by differentiating numerator and denominator by x we get,

$L=\lim_{x\to \frac{\pi}{4}}\Big[\frac{\sin x-\cos x}{8(4x-\pi)}\Big]$   again $(\frac{0}{0})$ form.

Again applying L-Hospital rule we get,

$L=\lim_{x\to \frac{\pi}{4}}\frac{\sin x+\cos x}{32}$

$=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{32}$

$=\frac{\sqrt{2}}{32}$

That is,

$\lim_{x\to \frac{\pi}{4}}\Big[\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}\Big]=\frac{\sqrt{2}}{32}$