Finding $ \lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x} $ I'm kind of stuck on this problem, I could use a hint.
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}
$$
After some algebra, I get
$$
{\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)-\sqrt{x+2}+\sqrt{2x}}}
$$
EDIT above should be:
$$
\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)(\sqrt{x+2}+\sqrt{2x)}}
$$
I'm stuck at this point, any help would be greatly appreciated.
 A: \begin{align}
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}&=
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x(x-2)}\frac{\sqrt{x+2}+\sqrt{2x}}{\sqrt{x+2}+\sqrt{2x}} \\
&=\lim_{x \to 2} \frac{(x+2)-(2x)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac{-(x-2)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac{-1}{x(\sqrt{x+2}+\sqrt{2x})} \\
&=\frac {-1}{2(\sqrt{2+2}+\sqrt{2(2)})} \\
&=\boxed{-\frac 18}
\end{align}
A: rewrite it in the form
$$\frac{(\sqrt{x+2}-\sqrt{2x})(\sqrt{x+2}+\sqrt{2x})}{x(x-2)(\sqrt{x+2}+\sqrt{2x})}$$
A: You can notice that the denominator is $x(x-2)$, you have
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}=
\left(\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}\right)
\left(\lim_{x\to 2}\frac{1}{x}\right)
$$
provided the limit exists. Thus we can concentrate on
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}
$$
which is the derivative at $2$ of the function
$$
f(x)=\sqrt{x+2}-\sqrt{2x}
$$
Since
$$
f'(x)=\frac{1}{2\sqrt{x+2}}-\frac{1}{\sqrt{2x}}
$$
we have
$$
f'(2)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}
$$
Thus your limit is
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}=
\left(\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}\right)
\left(\lim_{x\to 2}\frac{1}{x}\right)=
\left(-\frac{1}{4}\right)\left(\frac{1}{2}\right)=-\frac{1}{8}
$$
