# Limits at a point

• Jun 10th 2010, 12:12 AM
Glitch
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.
• Jun 10th 2010, 12:27 AM
Prove It
Quote:

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.
• Jun 10th 2010, 12:30 AM
Glitch
Ahh, I see. I was thinking that that'd be 0/0 (when you sub in 2) which is undefined. Thanks.