Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$ 
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What method to use? 
Result should be $\frac{1}{4}$

L'Hospital's rule is not the α and ω of limits computation. First remove the square roots in the numerator:
$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{(\sqrt{1+\tan x}+\sqrt{1+x})\sin^2x}$$
Now use equivalents:


*

*$\tan x-x=x+\dfrac{x^3}3+o(x^3)-x$, hence $\;\tan x-x\sim_0 \dfrac{x^3}3$

*$\sqrt{1+\tan x}+\sqrt{1+x}\xrightarrow[x\to 0]{}2$

*$\sin x\sim_0 x$


$$\text{So we have:}\hspace{8em}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}\sim_0\frac{\dfrac{x^3}3}{2x^2}=\frac x6\to 0.\hspace{8em}$$
