<s>[INST] 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.
 [/INST] Here's an approach :
Let :
$$L=\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}$$
\begin{align}
\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}&=\lim_{x\to 0}\frac{\bigg(\sqrt{1+x\sin x}-\sqrt{\cos x}\bigg)\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\lim_{x\to 0}\frac{1+x\sin x-\cos x}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\frac{1}2\Bigg(\lim_{x\to 0}\frac{x\sin x}{x\tan x}+\lim_{x\to 0}\frac{1-\cos x}{x\tan x}\Bigg)\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{\cos x(1-\cos x)}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{1-\cos x}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x(1-\cos x)}{x^2 \sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x}{\sin x}\frac{1-\cos x}{x^2}\\
&=\frac{1}2+\frac{1}4=\frac{3}4
\end{align}
Hence :
$$L=\frac{3}4$$
</s>