Question

The limiting value of the function $f\left(x\right)=\frac{2\sqrt{2}-(\cos x+\sin x{)}^3}{1-\sin 2x}$ when $x\to \frac{\pi }{4}$ is :

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

Answer 1

The correct option is C $3\sqrt{2}$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}-(cosx+sinx{)}^3}{1-sin2x}$

It is $\frac{0}{0}$ form

we solve it by L-Hospital Rule

$\underset{x\to \frac{\pi }{4}}{lim}\frac{0-3(cosx+sinx{)}^2(-sinx+cosx)}{0-2cos2x}$

again it form $\frac{0}{0}$ form

by L-hospital Rule

$\underset{x\to \frac{\pi }{4}}{lim}\frac{-3\times 2(cosx+sinx)(-sinx+cosx{)}^2-3(cosx+sinx{)}^2(-cosx-sinx)}{-4sin2x}$

put limit

$\frac{0-3(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}{)}^2(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})}{-4\left(1\right)}$

$=\frac{+3\left(2\sqrt{2}\right)}{-4}$

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