Answer
$\frac{1}{4}$
Work Step by Step
$\lim\limits_{x \to 0}\frac{\sqrt {1+\tan x}-\sqrt {1+\sin x}}{x^{3}}$
= $\lim\limits_{x \to 0}\frac{\sqrt {1+\tan x}-\sqrt {1+\sin x}}{x^{3}}\times\frac{{\sqrt {1+\tan x}+\sqrt {1+\sin x}}}{{\sqrt {1+\tan x}+\sqrt {1+\sin x}}}$
= $\lim\limits_{x \to 0}\frac{ {1+\tan x}- {1-\sin x}}{x^{3}[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}$
= $\lim\limits_{x \to 0}\frac{ {\tan x}-\sin x}{x^{3}[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}\times\frac{\cos x}{\cos x}$
= $\lim\limits_{x \to 0}\frac{\sin x(1-\cos x)}{x^{3}\cos x[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}$
= $\lim\limits_{x \to 0}\frac{\sin x(1-\cos x)}{x^{3}\cos x[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}\times\frac{1+\cos x}{1+\cos x}$
= $\lim\limits_{x \to 0}\frac{\sin x\sin^{2}x}{x^{3}\cos x(1+\cos x)[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}$
= $[\lim\limits_{x \to 0}\frac{\sin x}{x}]^{3}$$\lim\limits_{x \to 0}\frac{1}{\cos x(1+\cos x)[{\sqrt {1+\tan x}+\sqrt {1+\sin x}}]}$
=$1^{3}\times\frac{1}{(\sqrt 1+\sqrt 1)\times1\times(1+1)}$
= $\frac{1}{4}$
