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?

If you're willing to use a symbolic differentiation program like Matlab and have forgotten your trig and calculus formulas like I have, you should be able to recover the answer as $$\frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin x})}{dx^3}}{\frac{d^3(x^3)}{dx^3}} = \frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin x})}{dx^3}}{6}$$ evaluated at $x=0$.  This result is derived by applying the L'Hopital rule three times.  The denominator will reduce to $3!=6$, and the numerator can be calculated by the computer.  Even if you are trying to show your work, it never hurts to have a computer verify your answer.
