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.