Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$ 
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
And now:
$$\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x} = \infty$$
What I'm doing wrong? I know that answer is $\sqrt{3}$, but where is my mistake?
 A: $$\lim_{x\to0}\frac{x (\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}\color{red}x}{\color{red}2x} = \sqrt 3$$
Note the changes in red.
You didn't cancel correctly here.
A: Let $f(x) =\sqrt {3+x} - \sqrt {3-x}.$ Note $f(0)= 0.$ Our expression has the form
$$\frac{x-0}{f(x) - f(0)}.$$
By the definition of a derivative, the above $\to 1/f'(0)$ as $x\to 0.$ This is an easy computation.
A: How is $\;\;x(\sqrt{3+x}+\sqrt{3-x}) = 2\sqrt{3}x$?  Instead of simplifying any further you could also solve it by canceling and then taking the limit:
\begin{align}
\lim_{x\rightarrow 0} \frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\rightarrow 0} \frac{(\sqrt{3+x}+\sqrt{3-x})}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
\end{align}
A: $$
\underbrace{\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x}}_\text{I have no idea what you did here!}
$$
Let's recall some simple algebra:
$$
\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x}
= \overbrace{\frac{x(\cdots\cdots)}{x(\cdots\cdots)} = \frac{\sqrt{3+x}+\sqrt{3-x}}{2} \vphantom{\frac\int\int} }^\text{Cancel the $x$s.}
$$
Now you have
$$
\lim_{x\to0} \frac{\sqrt{3+x}+\sqrt{3-x}} 2 = \frac{2\sqrt 3} 2 = \sqrt 3.
$$
