Determine (if it exists) by evaluating the corresponding one-sided limits.

Determine (if it exists) by evaluating the corresponding one-sided limits.

lim                          | x-3 | 

x --> 3                 ---------------

                              x^2 -9

As x tends to 3 from above x-3 is positive so  |x-3|=x-3

so

$\displaystyle\lim_{x\rightarrow3^+}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{x-3}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{1}{2x}\\ =\frac{1}{6}$

BUT  as x tends to 3 from below x-3 is negative so  |x-3|=3-x

$\displaystyle\lim_{x\rightarrow3^-}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{3-x}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{-1}{2x}\\ =\frac{-1}{6}$

This means that the graph is not continuous at x=3 so the limit does not exist.

Here is a picture of the graph.