Question

If the value of $\underset{x\to 0}{lim}(2-\cos x\sqrt{\cos 2x}{)}^{\left(\frac{x+2}{x^2}\right)}$ is equal to $e^a$, then $a$ is equal to

Answer 1

$L=\underset{x\to 0}{lim}(2-\cos x\sqrt{\cos 2x}{)}^{\left(\frac{x+2}{x^2}\right)}$
of the form $1^{\infty }$
$L=e^{\underset{x\to 0}{lim}(2-\cos x\sqrt{\cos 2x}-1)\times \left(\frac{x+2}{x^2}\right)}\cdots \left(1\right)$
$L=e^{\underset{x\to 0}{lim}\frac{(1-\cos x\sqrt{\cos 2x})}{x^2}\times \left(x+2\right)}$
Now,
$\underset{x\to 0}{lim}\frac{(1-\cos x\sqrt{\cos 2x})}{x^2}$
By L' hospital rule
$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x\sqrt{\cos 2x}-\cos x\times \frac{1}{2\sqrt{\cos 2x}}\times (-2\sin 2x)}{2x}$
$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x\cos 2x+\sin 2x\cos x}{2x\sqrt{\cos 2x}}$
$\Rightarrow \underset{x\to 0}{lim}\left(\frac{\sin x\cos 2x}{2x\sqrt{\cos 2x}}\right)+\left(\frac{\sin 2x\cos x}{2x\sqrt{\cos 2x}}\right)$
$\Rightarrow \frac{1}{2}+1=\frac{3}{2}$
putting in equation $\left(1\right)$
$L=e^{\underset{x\to 0}{lim}\frac{3}{2}\times (x+2)}$
$L=e^{\frac{3}{2}\times 2}=e^3$
$\therefore$ $a=3$