Calculate the limit $\lim_{x\to0^+}\frac{\sqrt{\cos(2x)}-\sqrt{1+x\sin(x)}}{\tan^2\frac x2}$

Question

$$\lim_{x\to0^+}\frac{\sqrt{\cos(2x)}-\sqrt{1+x\sin(x)}}{\tan^2\frac x2}$$

I need to calculate this limit but I don't know what to do for to get out this indeterminate.

I make:

1. $\sqrt{\cos(2x)}-\sqrt{1+x\sin(x)} = \sqrt{\cos(2*0)}-\sqrt{1+0\sin(0)} = \sqrt{\cos(0)}-\sqrt{1+x\sin(0)} = \sqrt{1}-\sqrt{1} = 0 $

2. $\tan^2\frac{x}{2} = \frac{0}{2} = 0$

then, $\frac{0}{0}$

I want to get out the $tan^2\frac{x}{2}$.

Answer 1

$$ \lim_{x\to 0^+}\frac{\sqrt{\cos(2x)}-\sqrt{1+x\sin(x)}}{\tan^2\frac x2}\\[2em] = \lim_{x\to 0^+}\frac{\sqrt{\cos(2x)}-\sqrt{1+x\sin(x)}}{\tan^2\frac x2} \frac{\sqrt{\cos(2x)}+\sqrt{1+x\sin(x)}}{\sqrt{\cos(2x)}+\sqrt{1+x\sin(x)}}\\[2em] =\lim_{x\to 0^+} \frac{1}{\sqrt{\cos(2x)}+\sqrt{1+x\sin(x)}}\ \lim_{x\to 0^+}\frac{\cos(2x)-1-x\sin(x)}{\tan^2\frac x2}\\[2em] =\frac{1}{2}\ \lim_{x\to 0^+}\frac{\cos(2x)-1-x\sin(x)}{\tan^2\frac x2} $$ provided the remaining limit exists, but it can be handled using l'hopital or $\frac{\sin x}{x} \to 1$.