# Thread: Limits at a point

1. ## Limits at a point

Hey guys,

I'm trying to understand how to determine the left and right side of this limit:

limit as x -> 2

|x-2|/(x-2)

Do I have to graph it, or is there an algebraic way to do this? Cheers.

2. Originally Posted by Glitch
Hey guys,

I'm trying to understand how to determine the left and right side of this limit:

limit as x -> 2

|x-2|/(x-2)

Do I have to graph it, or is there an algebraic way to do this? Cheers.
Recall that

$\displaystyle |x - 2| = \begin{cases}\phantom{-(}x-2\phantom{)}\textrm{ if }x-2\geq 0\\-(x - 2)\textrm{ if }x-2<0\end{cases}$

$\displaystyle |x - 2| = \begin{cases}x-2\textrm{ if }x\geq 2\\2-x\textrm{ if }x<2\end{cases}$

When you are are approaching $\displaystyle x = 2$ from the left, $\displaystyle x < 2$.

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

$\displaystyle = \lim_{x\to 2^{-}}\frac{-(x-2)}{x-2}$

$\displaystyle = \lim_{x\to 2^{-}}(-1)$

$\displaystyle = -1$.

When you are approaching $\displaystyle x=2$ from the right, $\displaystyle x>2$.

So $\displaystyle \lim_{x \to 2^{+}}\frac{|x-2|}{x-2} = \lim_{x\to 2^{+}}\frac{x-2}{x-2}$

$\displaystyle =\lim_{x \to 2^{+}}(1)$

$\displaystyle = 1$.

Clearly, since the left hand limit is not the same as the right hand limit, the limit does not exist.

3. Ahh, I see. I was thinking that that'd be 0/0 (when you sub in 2) which is undefined. Thanks.