Consider the function:
[tex]h(x)= \frac{x^2-1}{x-1} [/tex]
As x tends to 1,
[tex]h(1)= \frac{1^2-1}{1-1} = \frac{0}{0} [/tex]
hence, the y value does not exist.
But
[tex] \lim_{x\to1} h(x)= \lim_{x\to1} \frac{x^2-1}{x-1} \\ \\ = \lim_{x\to1} \frac{(x-1)(x+1)}{x-1}= \lim_{x\to1} (x+1) \\ \\ =1+1=2[/tex]
Therefore, the limit of h(x) exists but the y-value does not exist.
