If we simply substitute the value x=2 in the expression we have
which is undefined.
But if we factor both numerator and denominator, we have
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
And we can evaluate this at 2 with no problems:
So, we have
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: