If we simply substitute the value x=2 in the expression we have
[tex]f(2) = \dfrac{4-12+8}{4-2-2} = \dfrac{0}{0}[/tex]
which is undefined.
But if we factor both numerator and denominator, we have
[tex]f(x)=\dfrac{x^2-6x+8}{x^2-x-2} = \dfrac{(x-2)(x-4)}{(x+1)(x-2)}[/tex]
Since we are studying the limit as x approaches 2, we can assume that x is not 2. In this case, we can simplify the (x-2) parenthesis, and the expression becomes
[tex]f(x)=\dfrac{x-4}{x+1}[/tex]
And we can evaluate this at 2 with no problems:
[tex]f(2) = \dfrac{2-4}{2+1} = -\dfrac{2}{3}[/tex]
So, we have
[tex]\displaystyle \lim_{x\to 2}f(x) = -\dfrac{2}{3}[/tex]
This means that in this case both left and right limits exist and are the same, so the limit exists, but the function is not defined at x=2. This is a removable discontinuity, because we can define the function as its limit, and we have a continuous function at x=2:
[tex]f(x) = \begin{cases}\dfrac{x^2-6x+8}{x^2-x-2} &\text{if }x \neq 2\\ -\frac{2}{3} &\text{if }x = 2\end{cases}[/tex]
