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?

Solution 1:

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.

Solution 2:

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}

Solution 3:

We have $\sqrt{x^3+4x^2}=\sqrt{x^2(x+4)}=|x|\sqrt{x+4}$ !

Now cosider two cases:

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

Solution 4:

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 $x <0$.

Solution 5:

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 $x \rightarrow 0^{\pm}$.