Question

The value of $\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}-{(\cos x+\sin x)}^3}{1-\sin 2x}$ is

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{\sqrt{2}}{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\frac{3}{\sqrt{2}}$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}-{(\cos x+\sin x)}^3}{1-\sin 2x}$ $\left(\frac{0}{0}orm\right)$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-3{(\cos x+\sin x)}^2(-\sin x+\cos x)}{-2\cos 2x}$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-3(\cos x+\sin x)({\cos }^{2}x-{\sin }^{2}x)}{-2\cos 2x}$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-3(\cos x+\sin x)\cos 2x}{-2\cos 2x}$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{3(\cos x+\sin x)}{2}=\frac{3}{2}.(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=\frac{3}{\sqrt{2}}$