<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] $$\lim_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=L$$
Using L'hopital: 
$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{2\sin x \cos x}=L$$
Reordering the denominator:
$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{\sin 2x }=L$$
Using L'hopital Again
$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x }=L$$
You cannot use l'hopital again because $\lim_{x\to 0} \cos 2x = 1 > 0$
so replacing:
$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x } = \dfrac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2\cdot 1} = 0$$
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