<s>[INST] Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate 
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
factor out $\sin x$ in the numerator
$$\lim_{x \to 0} \frac{\sin x \cdot (\sec x - 1)}{x^3 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
simplify using $\lim_{x \to 0} \frac{\sin x}{x} = 1 $
$$\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}$$
From here I don't see any useful direction to go in, if I even went in an useful direction in the first place, I don't know.

I suspect that this could be evaluated using the definition of derivatives, if so, or not, any suggestions?
 [/INST] $\begin{align}
\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} &= \lim_{x\to 0}\frac{1}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}\cdot\lim_{x\to 0}\frac{\sec x-1}{x^2}\\
&=\frac12 \cdot \lim_{x\to 0}\frac{\sec x-1}{x^2}
\end{align}$
This last limit should yield to L'Hopital's rule, or you could multiply by the conjugate of the numerator and apply a trig identity, plus the fact that $\frac{\tan x}{x}\to 0$ as $x\to 0$.
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