How to find $\lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x}$ when $x\to 0^+$ and when $x\to 0^-$? I'm trying to find:
$$ \lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x} $$
Since there is a discontinuity at $x=0$ I know that I have to take the limits from both sides, $x \to 0^+$ and $x \to 0^-$, and check if they're equal.

If I factor it I get:
$$ \lim \limits_{x \to 0} \left(\frac{\sqrt{x+4}} {x-1}\right) = - 2$$
Is this the same as $x \to 0^+$?
If so, how do I approach the problem for $x \to 0^-$?
If not, how do I do I do it from both sides? 

Note : $ \sqrt{x^3+4x^2}=|x|\sqrt{x+4}$.
We have  $\dfrac{|x|\sqrt{x+4}}{x(x-1)}$.
For $x>0$: $\dfrac{|x|}{x}=1$;
For $x <0$ $\dfrac{|x|}{x}=-1$;
Now proceed to take limits $x \rightarrow 0^{\pm}$.
