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.

We use the elementary limit results $\displaystyle  \lim_{x\to 0}\frac{\sin x}{x} =1=\lim_{x\to 0} \frac{\tan x}{x}$
Now coming to the main problem we may  write it as $$\begin{aligned}\lim_{x\to 0} \frac{\sqrt{1+x^2\left(\frac{\sin x}{x}\right)}-\sqrt{\cos x}}{x^2\left(\frac{\tan x}{x}\right)}&=\lim_{x\to 0} \frac{\sqrt{1+x^2}-\sqrt{\cos x}}{x^2}\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots -\sqrt{1-\underbrace {\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots}}_{q}\right)\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x^2}{2}+O(x^4)-\left(1-\frac{q}{2}+\frac{q^2}{4}+O(q^6)\right)\right)\\&=\lim_{x\to 0} \frac{1}{x^2}\left(1+\frac{x^2}{2}-1+\frac{q}{2}-O(q^4)\right) \\&=\lim_{x\to 0} \frac{1}{x^2}\left(\frac{x^2}{2}+\frac{1}{2}\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6)\right)\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\end{aligned}$$
