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.

$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$
$=\lim_{x \to 0} \frac{(\sqrt{1 + x\sin x} - \sqrt{\cos x})(\sqrt{1 + x\sin x} + \sqrt{\cos x})}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{1+x\sin x-\cos x}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{x\sin x+2\sin^2 {x\over 2}}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
divided by $x^2$
$$=\lim_{x \to 0} \frac{\frac{\sin x}{x}+{1\over 2}\frac{(\sin {x\over 2})^2}{(x/2)^2}}{\frac{\tan x}{x}(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
$$=\frac{1+\frac12}{1(1+1)}$$
$$=\frac34$$
