<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] $  \lim_{x\to 0} \frac { \sqrt{1+\tan(x)} - \sqrt{1+sin(x)}} { x^3 } = \lim_{x\to 0} \frac {tan(x) - sin(x)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} =  \lim_{x\to 0} \frac {sin(x)(\frac {1} {cos(x)} - 1)} {x^3(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {1 - cos(x)} {cos(x)x^2(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \lim_{x\to 0} \frac {x^2} {2cos(x)x^2(\sqrt{1+\tan(x)} + \sqrt{1+sin(x)})} = \frac {1} {4} $
Intermediate calculations:
$$ \lim_{x\to0} \frac {2(1 - cos(x))} {x^2} = 1 $$
$$ \lim_{x\to0} cos(x) = 1 $$
$$ \lim_{x\to0} \frac {sin(x)}{x} = 1 $$
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