Classify the discontinuity at x = 3

For $x \neq 3$, factor the numerator: $\frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3$. Since $x \neq 3$ in this piece, the cancellation is valid, so $d(x) = x+3$ for all $x$ near 3 except at 3 itself. Therefore $\lim_{x\to 3} d(x) = \lim_{x\to 3}(x+3) = 6$.

Checking the three conditions for continuity at $x=3$: (1) $d(3) = 5$ is defined, (2) $\lim_{x\to 3} d(x) = 6$ exists, but (3) the limit does not equal $d(3)$ since $6 \neq 5$. The third condition fails, so $d(x)$ is discontinuous at $x=3$. Since the limit exists but does not equal the function value, this is a removable discontinuity. The discontinuity could be removed by redefining $d(3) = 6$.