<s>[INST] 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 $\frac{1}{4}$
 [/INST] If you know that $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0\tag{1}$$ then it is easy to give a simple evaluation for the limit in question
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{2}\{\sqrt{1 + \tan x} + \sqrt{1 + x}\}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\notag\\
&= 0\notag
\end{align}
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