Checking the three-part continuity test at x = 2

First check condition 1: $f(2)$ is given directly in the definition as $f(2) = 3$, so this value is defined. Next, for $x \neq 2$, simplify $f(x) = \dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$, since $x-2 \neq 0$ allows cancellation. This simplification lets us compute the limit using a polynomial expression instead of the original rational expression.

Now check condition 2: $\lim_{x\to 2} f(x) = \lim_{x \to 2}(x+2) = 4$, so the limit exists and equals $4$. Finally check condition 3: compare the limit value $4$ to $f(2) = 3$ — they are not equal, so condition 3 fails. Therefore $f$ is not continuous at $x=2$. Since the limit exists but does not equal $f(2)$, this is a removable discontinuity.