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.

Similar to Varun Vejalla's answer, but without L'Hopital: Multiplying by the conjugate and replacing $\tan x$ by $\sin x/\cos x$, we have
$$\begin{align}
{\sqrt{1+x\sin x}-\sqrt\cos x\over x\tan x}
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\cdot{1+x\sin x-\cos x\over x\sin x}\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos x\over x\sin x}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos^2x\over x\sin x(1+\cos x)}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{\sin x\over x}\cdot{1\over1+\cos x} \right)\\
&\to{1\over\sqrt{1+0}+\sqrt1}\left(1+1\cdot{1\over1+1}\right)={1\over2}\left(1+{1\over2}\right)={3\over4}
\end{align}$$
Note, we also multiplied by the conjugate of $1-\cos x$, made use of the trig identity $1-\cos^2x=\sin^2x$, and, finally, assumed the familiar limit ${\sin x\over x}\to1$ as $x\to0$.
