Left-hand limit at x = 2

Since we are approaching $x=2$ from the left, we consider values of $x$ slightly less than 2, such as $1.9, 1.99, 1.999$. For all such values, $x < 2$, so we must use the rule $f(x) = 3x+1$.

As $x \to 2^-$, we substitute into $3x+1$: the values approach $3(2)+1 = 7$. Therefore $\lim_{x\to 2^-} f(x) = 7$.

Since we are approaching $x=2$ from the right, we consider values of $x$ slightly greater than 2, such as $2.1, 2.01, 2.001$. For all such values, $x \geq 2$, so we must use the rule $f(x) = x^2+3$.

As $x \to 2^+$, we substitute into $x^2+3$: the values approach $(2)^2+3 = 7$. Wait — let's recompute carefully: $2^2 + 3 = 4+3 = 7$. So $\lim_{x\to 2^+} f(x) = 7$.

From part (a), $\lim_{x\to 2^-} f(x) = 7$, and from part (b), $\lim_{x\to 2^+} f(x) = 7$. Since both one-sided limits exist and are equal to the same value.

the two-sided limit exists and equals that common value: $\lim_{x\to 2} f(x) = 7$.