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

Factor the numerator: $x^2 + x - 6 = (x-2)(x+3)$. So for $x \neq 2$, $g(x) = \frac{(x-2)(x+3)}{x-2} = x+3.$ This cancellation is only valid where $x \neq 2$, since $g(2)$ is undefined (division by zero).

Since $\lim_{x\to 2} g(x) = \lim_{x\to 2}(x+3) = 5$ exists but $g(2)$ is not defined, the limit exists but is not equal to $g(a)$ (in fact $g(a)$ doesn't even exist). This means the discontinuity is removable, and the hole is located at the point $(2, 5)$.

Yes. The earlier work showed the limit of g as x approaches two equals five, but g of two was originally undefined. Since a removable discontinuity occurs exactly when the limit exists but doesn't match the function's value (or the value is missing), setting g of two equal to five fills the hole and satisfies all three continuity conditions there.