Question

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

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

Answer 1

Answer: (A)

$\frac{3}{\sqrt{2}}$

$\underset{x\to \pi /4}{lim}\frac{2\sqrt{2}-(\cos x+\sin x{)}^3}{1-\sin 2x}$
Applying L-Hospital's rule, we get
$=\underset{x\to \pi /4}{lim}\frac{-3(\cos x+\sin x{)}^2(\cos x-\sin x)}{-2\cos 2x}$

$=\underset{x\to \pi /4}{lim}\frac{-3\cos 2x(\cos x+\sin x)}{-2\cos 2x}$

$=\frac{3}{\sqrt{2}}$

Hence, option 'A' is correct.