How to find $\lim\limits_{x \to 0} \frac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$? How to find $\;\;\lim\limits_{x \to 0} \dfrac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$ ?
 A: Hint: Multiply the numerator and denominator by $\sqrt{1 + \tan x} + \sqrt {1-\tan x}$.
Your denominator will become $1 + \tan x - (1-\tan x) = 2\tan x$. 
A: We have
$$f(x) = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} \dfrac{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}$$
Hence,
$$f(x) = \dfrac{\sin(2x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{1+\tan(x) - 1 + \tan(x)} = \dfrac{2\sin(x)\cos(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{2\tan(x)}$$
This gives us
$$f(x) = \cos^2(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)$$
Now take the limit as $x \to 0$ to get $f(x) \to 2$.
A: Sinc you received already good answers, let me show you what you can do using Taylor series. You know that, for small values of $x$, $\tan(x)\approx x$ and that $\sin(x)\approx x$. So the expression is almost $$\frac{2x}{\sqrt{1+x}-\sqrt{1-x}}$$ Now remember that $\sqrt{1+y}\approx \frac y2$. So the expression is $$\frac{2x}{(\frac x2)-(-\frac x2)}$$
