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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?

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

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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}

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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. $x \to 0^{-}$.
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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 $x <0$.

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

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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} $

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