# 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?

• For $x \to 0^{-}$ the limit is $2$. Put $y=-x$ and take limit as $y \to 0^{+}$ – Kavi Rama Murthy Feb 13 at 10:31

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:

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

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 $$x <0$$.

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

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

$$\lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\ = \lim \limits_{x \to 0^-} \left(\frac{ |x| \sqrt{x+4}} { x(x-1) }\right) \\ = \lim \limits_{\delta \to 0} \left(\frac{ |0-\delta| \sqrt{ (0-\delta) +4}}{ (0-\delta)( (0-\delta) -1 ) }\right) \ [ \ \text{substituting} \ x = 0 - \delta \ , \delta > 0 \ ] \\ = \lim \limits_{\delta \to 0} \left(\frac{ -\delta \sqrt{ 4 - \delta }}{ \delta (-1 - \delta) }\right) \\ = 2$$

$$\therefore \ \lim \limits_{x \to 0^+} \frac{\sqrt{x^3+4x^2}} {x^2-x} \neq \lim \limits_{x \to 0^-} \frac{\sqrt{x^3+4x^2}} {x^2-x} \\ \Rightarrow \lim \limits_{x \to 0} \frac{\sqrt{x^3+4x^2}} {x^2-x} \ \text{does not exist}$$