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? 
 A: Note that the answer depends on the sign of $x$:
\begin{align}
\frac{\sqrt{x^3+4x^2}} {x^2-x}
&=\frac{2|x|\sqrt{1+\frac x4}}{-x(1-x)}\\
&=\begin{cases}
-2\frac{\sqrt{1+x/4}}{1-x}&x\to 0^+\\
2\frac{\sqrt{1+x/4}}{1-x}&x\to 0^-\\
\end{cases}
\end{align}
A: We have $\sqrt{x^3+4x^2}=\sqrt{x^2(x+4)}=|x|\sqrt{x+4}$ !
Now cosider two cases:


*

*$x \to 0^{+}$ and 2. $x \to 0^{-}$.

A: Hint: if you factor, you get 
$$\frac{|x| \sqrt{x+4}}{x(x-1)} $$
Consider that $|x|/x = 1$ if $x > 0$ and $|x| / x = -1$ if $x <0$.
A: Limit from right side is
$
\lim \limits_{x \to 0^+} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
= \lim \limits_{x \to 0^+} \left(\frac{ |x| \sqrt{x+4}} { x(x-1) }\right) \\
= \lim \limits_{\delta \to 0} \left(\frac{ |0+\delta| \sqrt{ (0+\delta) +4}}{ (0+\delta)( (0+\delta) -1 ) }\right) \ [ \ \text{substituting} \ x = 0 + \delta \ , \delta > 0 \ ] \\
= \lim \limits_{\delta \to 0} \left(\frac{ \delta \sqrt{ \delta+4}}{ \delta (\delta-1) }\right) \\
= -2
$ 
Limit from left side is
$
\lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
= \lim \limits_{x \to 0^-} \left(\frac{ |x| \sqrt{x+4}} { x(x-1) }\right) \\
= \lim \limits_{\delta \to 0} \left(\frac{ |0-\delta| \sqrt{ (0-\delta) +4}}{ (0-\delta)( (0-\delta) -1 ) }\right) \ [ \ \text{substituting} \ x = 0 - \delta \ , \delta > 0 \ ] \\
= \lim \limits_{\delta \to 0} \left(\frac{ -\delta \sqrt{ 4 - \delta }}{ \delta (-1 - \delta) }\right) \\
= 2
$
$
\therefore \ 
\lim \limits_{x \to 0^+} \frac{\sqrt{x^3+4x^2}} {x^2-x}
\neq \lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\
\Rightarrow \lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x} \ \text{does not exist}
$ 
A: 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 $x \rightarrow 0^{\pm}$.
A: Note that your expression after factoring, becomes :
$$\frac{\sqrt{x^3 + 4x^2}}{x^2-x} = \frac{\sqrt{x^2(x+4)}}{x(x-1)} = \frac{|x|\sqrt{x+4}}{x(x-1)}$$
This is exactly where your mistake is. When you factor under the square root, $x^2$ becomes $|x|$. That means, by definition of the absolute value, that :
$$|x| = \begin{cases} x &x\geq 0 \\-x &x<0 \end{cases}$$
Eventually, the left sided limit will be $2$ and the right sided $-2$, which means that the total limit does not exist.
