Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.

We can work asymptotically also:
We denote the limit by $L$
As $x \to 0$
$$L=\frac{(\sqrt{1+\tan x}-1) -(\sqrt{1+\sin x}-1)}{x^3} \sim \frac{1}{2} \,\,\left[ \frac{\tan x (1-\cos x)}{x^3} \right] \sim \frac{1}{4}\,\,\frac{\tan x}{x} \sim \frac{1}{4}$$
