How to find $\lim_{x\to 2}\frac{|x-2|}{2x-x^2}$ $$\lim_{x\to 2}\frac{|x-2|}{2x-x^2}$$
I know the answer of the left hand limit is $1/2$; while the right hand limit is $-1/2$. But I don't understand how do you get that?
If I factor $-x$ from the denominator, I'll get $(-2+x)$ which cancels out with the numerator. Then I'll get $1/-x$. If I plug in the limit of $2$ from the left hand, it would be $1/2$. Wouldn't it also be $1/2$ from the right hand, as well? I'm getting confused on how to work this out. Please help, thank you!
 A: The best way is to consider separately the two cases

*

*$x>2$ that is $x \to 2^+$ we have $|x-2|=x-2$ and thereofore

$$\lim_{x\to 2^+}\frac{|x-2|}{2x-x^2}=\lim_{x\to 2^+}\frac{x-2}{x(2-x)}=\lim_{x\to 2^+}-\frac{1}{x}$$

*

*$x<2$ that is $x \to 2^-$ we have $|x-2|=-x+2$ and thereofore

$$\lim_{x\to 2^-}\frac{|x-2|}{2x-x^2}=\lim_{x\to 2^-}\frac{-x+2}{x(2-x)}=\lim_{x\to 2^-}\frac{1}{x}$$
A: You might find it easier to analyze with the substitution $u = x-2$. Then it becomes
$\lim_{u \rightarrow 0}\frac{|u|}{-u(u+2)}=\lim_{u \rightarrow 0}\frac{-1}{u+2}\frac{|u|}{u}=\left(\lim_{u \rightarrow 0}\frac{-1}{u+2}\right)\left( \lim_{u \rightarrow 0}\frac{|u|}{u}\right)=\frac {-1}2 \left(\lim_{u \rightarrow 0}\frac{|u|}{u}\right) $
$\frac{|u|}u$ is just $sign(u)$: it's $-1$ for $u<0$ and $1$ for $u>0$.
A: first:
if $f(x) = \frac{|x-2|}{2x-x^2}$
then :
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{x-2}{x\cdot (2-x)} = \lim_{x \to 2^+} -\frac{1}{x} = -\frac{1}{2}$
and
$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2-x}{x\cdot (2-x)} = \lim_{x \to 2^-} \frac{1}{x} = \frac{1}{2}$
but from the definition we have $\lim_{x \to 2} f(x) =l \Leftrightarrow \lim_{x \to 2^+} f(x) =\lim_{x \to 2^-} f(x) =l$ but $\frac{1}{2} \neq -\frac{1}{2}$ thus limit of $f(x)$ at $x=2$ does not exist.
A: Piece-wise definition of a modulus states that
whenever
$$ a>0 => |a| = a$$
And $$a<0 => |a| = -a$$
Because the modulus of anything should be positive.
If taking limit from LHS, $$\lim_{x\to 2^-} x<2 => x-2<0 => |x-2|= -(x-2)$$
$$\lim_{x\to 2^-}\frac{|x-2|}{2x-x^2} = \frac{-(x-2)}{x(2-x)}= \frac{1}{2}$$
Similarly if $$x>2=>x-2>0 => |x-2| = x-2$$
Therefore the limit is  negative half.
