<s>[INST] 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.
 [/INST] \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}
</s>