Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.

After multiplying the numerator and denominator of the expression in the limit by $$\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}$$, I get $$\lim_{x \to 0} \frac{1+x\sin(x)-\cos(x)}{x\tan(x) \left(\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}\right)}$$
It is clear that $$\lim_{x \to 0}\left( \sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)} \right) = \sqrt{1 + 0 \sin(0)} + \sqrt{\cos(0)} = 2$$
so the original limit simplifies to $$\frac{1}{2} \lim_{x \to 0} \frac{1 + x\sin(x) - \cos(x)}{x \tan(x)} = \frac{1}{2} \left( \lim_{x \to 0} \frac{x\sin(x)}{x\tan(x)} + \lim_{x \to 0} \frac{1-\cos(x)}{x \tan(x)} \right)$$
The first limit is simply $\cos(0) = 1$, so the limit becomes $$\frac{1}{2} \left(1 +  \lim_{x \to 0}\frac{1 - \cos(x)}{x \tan(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{\cos(x) (1 - \cos(x))}{x \sin(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} \right)$$
Then using L'Hôpital's rule, I get that $$\lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x\cos(x) + \sin(x)}$$
Using L'Hôpital's rule once more: $$\lim_{x \to 0} \frac{\cos(x)}{2\cos(x) - x\sin(x)} = \frac{1}{2}$$
and therefore $$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} = \frac{1}{2} \left(1 + \frac{1}{2} \right) = \frac{3}{4}$$
