Checking continuity at x = 2 using the three-part test

First check that $g(2)$ is defined: since $2 \ge 2$, we use $g(x) = x^2+3$, so $g(2) = 2^2+3 = 7$. Next find the left-hand limit using the piece for $x<2$: $\lim_{x\to 2^-} g(x) = \lim_{x\to 2^-}(3x+1) = 3(2)+1 = 7$. Then find the right-hand limit using the piece for $x\ge 2$: $\lim_{x\to 2^+} g(x) = \lim_{x\to 2^+}(x^2+3) = 2^2+3 = 7$.

Since the left-hand and right-hand limits both equal $7$, the two-sided limit $\lim_{x\to 2} g(x) = 7$ exists. Comparing this to $g(2) = 7$, all three conditions of continuity are satisfied: $g(2)$ is defined, the limit exists, and the limit equals $g(2)$. Therefore $g$ is continuous at $x=2$, and there is no discontinuity to classify.