Shorter way to calculate this limit I want to calculate this (without L'Hospital) :
$$\lim_{x \to 0} \frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3}$$
I already solved it in a looooong way:
\begin{align}
\frac{\sqrt{1+ \tan x} - \sqrt{1+ \sin x}}{x^3} &= \frac{\tan x - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
\frac{\sin x (1- \cos x )}{x^3 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
\frac{\sin x}{x} \frac{(1- \cos x)(1+\cos x) }{x^2 \cos x \ (1+\cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
\frac{\sin x}{x} \frac{1- \cos^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
\frac{\sin x}{x} \frac{\sin^2 x }{x^2} \frac{1}{\cos x \ (1 + \cos x)} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}  \\ \\ &=
1 \cdot 1^2 \cdot \frac{1}{1 \cdot (1 +1)} \frac{1}{\sqrt{1+0}+\sqrt{1+0}} \tag{as $x \to 0$} \\ \\ &=
\frac 1 4
\end{align}
Is there a shorter way?
EDIT: I am wondering whether there is a trick like $x= \arctan u^4$ etc.

Your proof looks correct (but you slightly abused notation at the end - you should say $\to 1 \cdot 1^2 \cdots$, not $= 1 \cdot 1^2 \cdots$). You can shorten it if you use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.
$$\frac{\sin x}{x} \frac{1- \cos x }{x^2 \cos x} \frac{1}{\sqrt{1+ \tan x} + \sqrt{1+ \sin x}}$$
$$=\frac{\sin x}{x}\frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x\left(\sqrt{1+\tan x}+\sqrt{1+\sin x}\right)}$$
$$\stackrel{x\to 0}\to 1\cdot \frac{1}{2}\cdot \frac{1}{1\cdot (\sqrt{1+0}+\sqrt{1+0})}=\frac{1}{4}$$
