How to find $\lim_{x\to 2}\frac{|x-2|}{2x-x^2}$

Question

$$\lim_{x\to 2}\frac{|x-2|}{2x-x^2}$$

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!

Answer 1

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

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

Answer 2

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.

Answer 3

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 $u>0$.

Answer 4

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.