Question

The value of $\underset{x\to \frac{x}{4}}{lim}\frac{4\sqrt{2}-{(\cos x+\sin x)}^5}{1-\sin 2x}$ is

• A. $0$
• B. $\sqrt{2}$
• C. $5\sqrt{2}$
• D. $3$

Answer 1

The correct option is C $5\sqrt{2}$

$Given,{lim}_{x\to \frac{x}{4}}\frac{4\sqrt{2}-{\left(\cos x+\sin x\right)}^5}{1-\sin 2x}$

Apply L-hospital rule,

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{\frac{d}{dx}\left(4\sqrt{2}\right)-\frac{d}{dx}{\left(\cos x+\sin x\right)}^5}{\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(\sin x\right)}$

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{0-5{\left(\cos x+\sin x\right)}^4\left\{-\sin x+\cos x\right\}}{-2\cos 2x}$

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{5{\left(\cos x+\sin x\right)}^4\left\{\cos x-\sin x\right\}}{2\cos 2x}$

Again apply L-hospital rule,

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{5\left\{4{\left(\cos x+\sin x\right)}^3{\left(\cos x-\sin x\right)}^2+\left(-\sin x-\cos x\right){\left(\cos x+\sin x\right)}^4\right\}}{-4\sin 2x}$

$\Rightarrow {lim}_{x\to \frac{x}{4}}\frac{5\left\{4{\left(\cos x+\sin x\right)}^3{\left(\cos x-\sin x\right)}^2-{\left(\cos x+\sin x\right)}^5\right\}}{-4\sin 2x}$

On putting the limits, we get

$=\frac{5\left\{0-{\left(\frac{2}{\sqrt{2}}\right)}^5\right\}}{-4\times 1}=\frac{5}{4}\times 4\sqrt{2}=5\sqrt{2}$

Hence, this is the answer.