Three-part continuity test at x = 3

First check that $u(3)$ is defined: by the definition of the piecewise function, $u(3) = 5$, so condition 1 is satisfied. Next, to find $\lim_{x\to 3} u(x)$, factor the numerator: $x^2 - x - 6 = (x-3)(x+2)$. For $x \neq 3$, we can cancel the common factor $(x-3)$ in $\dfrac{(x-3)(x+2)}{x-3}$, giving $u(x) = x + 2$ near $x=3$ (but not at $x=3$).

So $\lim_{x \to 3} u(x) = \lim_{x\to 3}(x+2) = 5$, meaning condition 2 (the limit exists) is satisfied and it equals $5$. Now compare: $\lim_{x\to3} u(x) = 5$ and $u(3) = 5$, so condition 3, $\lim_{x\to3}u(x) = u(3)$, also holds. All three conditions are met, so $u(x)$ is continuous at $x=3$.

No, u would no longer be continuous at three. From part a, the limit of u as x approaches three equals five. If u of three were redefined as four, the third condition would fail because the limit would not equal the function value. This would create a removable discontinuity, since the limit still exists but disagrees with the function's value at that point.