Classify the discontinuity of t(x) at x = 2

For $x \neq 2$, factor the numerator: $\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$. So as $x \to 2$, $t(x) \to 2+2 = 4$, meaning $\lim_{x\to 2} t(x) = 4$. This limit exists and is finite.

The function value is $t(2) = 5$ by definition. Since the limit exists (equal to 4) but does not equal $t(2) = 5$, one of the three continuity conditions fails while the other two hold. Because the limit exists but disagrees with the function's assigned value, the discontinuity is removable — the graph has a hole at $(2,4)$ that could be patched by redefining $t(2) = 4$.

The answer is four. Since the limit of t of x as x approaches two equals four, redefining t of two to be four would make the limit equal the function value. This satisfies all three conditions of continuity and removes the hole in the graph. This confirms the discontinuity was removable, since patching one point fixes it.