Question

${lim}_{x\to \frac{x}{4}}\frac{4\sqrt{2}-(cosx+sinx{)}^5}{1-sin2x}$ is equal to

• A. $5\sqrt{2}$
• B. $3\sqrt{2}$
• C. $\sqrt{2}$
  
• D. None of these

Answer 1

The correct option is A

$5\sqrt{2}$

${lim}_{x\to \frac{x}{4}}\frac{4\sqrt{2}-(cosx+sinx{)}^5}{1-sin2x}$
$={lim}_{x\to \frac{x}{4}}\frac{2^{\frac{5}{2}-{\left(\right(cosx+sinx{)}^2)}^{\frac{5}{2}}}}{2-(1+sin2x)}$
$={lim}_{x\to \frac{x}{4}}\frac{2^{\frac{5}{2}-{\left(\right(cosx+sinx{)}^2)}^{\frac{5}{2}}}}{2-(cosx+sinx{)}^2}$
$Lett=(cosx+sinx{)}^2$
$x\to \frac{\pi }{4}$
$\therefore t={(cos\frac{\pi }{4}+sin\frac{\pi }{4})}^2\to (\sqrt{2}{)}^2=2$
$={lim}_{t\to 2}\frac{2\frac{5}{2}-\left(t\right)\frac{5}{2}}{2-\left(t\right)}$
$=\frac{5}{2}(2{)}^{\frac{3}{2}}[\because {lim}_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}]$
$=5\sqrt{2}$