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 $\frac{1}{4}$
 A: 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 $\;\tan x-x\sim_0 \dfrac{x^3}3$

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

*$\sin x\sim_0 x$


$$\text{So we have:}\hspace{8em}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}\sim_0\frac{\dfrac{x^3}3}{2x^2}=\frac x6\to 0.\hspace{8em}$$
A: Using Taylor series we have
$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{\sin^2x(\sqrt{1+\tan x}+\sqrt{1+x})}\sim_0\frac{\frac13x^3}{2x^2}=\frac1{6}x$$
so the desired limit is $0$.
A: $$\lim_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=L$$
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:
$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x } = \dfrac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2\cdot 1} = 0$$
A: 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}
A: This can be (almost) completely solved using algebraic manipulation:
$$\frac{\sqrt{\tan x+1}-\sqrt{x+1}}{\sin^2(x)} = \frac{\tan x -x}{\sin^2 x\cdot(\sqrt{\tan x+1} +\sqrt{x+1})}$$
Now, since:
$$\lim_{x\to 0}\frac{1}{\sqrt{\tan x+1} +\sqrt{x+1}}=1$$
Our problem reduces to:
$$\lim_{x\to 0}\frac{\tan x -x}{\sin^2 x}= \lim_{x\to0}\frac{1}{\sin x}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg)=$$
$$\lim_{x\to0}\frac{1}{\sin x}\cdot\lim_{x\to0}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg)$$
Also,
$$\lim_{x\to0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\tan x}{\sin x} = 1$$
Then:
$$\lim_{x\to0}\frac{1}{\sin x}\cdot\lim_{x\to0}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg) = \lim_{x\to 0}\frac{1}{\sin x}\cdot 0 = 0$$
I am aware of the issue I have in my solution - most Calculus instructors I know would accept this work for this question.
