Checking the three-part continuity test at x = 2

By definition, $r(2) = 12$, so the first condition (f(a) is defined) is satisfied. To find the limit, factor the numerator using the difference of cubes formula: $x^3 - 8 = (x-2)(x^2+2x+4)$. For $x \neq 2$, this gives $r(x) = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$.

Taking the limit as $x \to 2$ of the simplified expression gives $\lim_{x\to 2} r(x) = 4 + 4 + 4 = 12$. Since the limit exists and equals $12$, all three conditions — $r(2)$ defined, the limit existing, and the limit equaling $r(2)$ — are satisfied. Therefore $r(x)$ is continuous at $x=2$, and there is no discontinuity to classify.