How to find $\lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x}$ when $x\to 0^+$ and when $x\to 0^-$?
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.
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}
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. $x \to 0^{-}$.