Question

$\underset{x\to \frac{\pi }{4}}{Lt}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi {)}^2}=$

• A. $\frac{1}{16\sqrt{2}}$
• B. $\frac{1}{32\sqrt{2}}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{16\sqrt{2}}$

$H_{x\to \pi /4}\frac{\sqrt{2}-cosx-sinx}{{(4x-\pi )}^2}=\frac{0}{0}$ form

Applying $L-$ Hospital's rule

$H_{x\to \pi /4}\frac{sinx-cosx}{2(4x-\pi )\left(4\right)}=\frac{0}{0}$ form

Again applying $L-$ Hospital rule

$H_{x\to \pi /4}\frac{cosx+sinx}{8\times 4}$

$=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{32}=\frac{\sqrt{2}}{32}=\frac{1}{16\sqrt{2}}$