Jtdylktuy

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?

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 :

$$|x| = begin{cases} x &xgeq 0 \-x &x<0 end{cases}$$

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}
-2frac{sqrt{1+x/4}}{1-x}&xto 0^+\
2frac{sqrt{1+x/4}}{1-x}&xto 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}
$