Answer
$\frac 14$
Work Step by Step
$\begin{aligned}&\lim\limits_{x\to 0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\\
=&\lim\limits_{x\to 0}\frac{1+\tan x-1-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\
=&\lim\limits_{x\to 0}\frac{\tan x -\sin x}{2x^3}\\
=&\lim\limits_{x\to 0}\frac{x+\frac 13 x^3-(x-\frac 16x^3)}{2x^3}\\
=&\lim\limits_{x\to 0}\frac 12\cdot \frac 12\\
=&\frac 14\end{aligned}$
