Question

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

• A. $5\sqrt{2}$
• B. $3\sqrt{2}$
• C. $\sqrt{2}$
• D. $7\sqrt{2}$

Answer 1

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

We have,

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

Apply L-Hospital rule,

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

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

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

$\Rightarrow {lim}_{x\to \frac{\pi }{4}}\left(\frac{20{(\cos x+\sin x)}^3(\cos x-\sin x)-5{(\cos x+\sin x)}^5}{-4\sin 2x}\right)$

$\Rightarrow \frac{20{(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})}^3(\cos \frac{\pi }{4}-\sin \frac{\pi }{4})-5{(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})}^5}{-4\sin \frac{\pi }{2}}$

$\Rightarrow \frac{0-5{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}^5}{-4}$

$\Rightarrow \frac{5{\left(\frac{2}{\sqrt{2}}\right)}^5}{4}$

$\Rightarrow \frac{5\left(\frac{32}{4\sqrt{2}}\right)}{4}$

$\Rightarrow \frac{40}{4\sqrt{2}}$

$\Rightarrow \frac{10}{\sqrt{2}}$

$\Rightarrow 5\sqrt{2}$

Hence, this is the answer.