Three-part continuity check at x = 3

First check $f(a)$ is defined: since $3\le 3$, we use the rule $m(x)=2^x$, so $m(3) = 2^3 = 8$. Next check the limit: the left-hand limit uses $2^x$ so $\lim_{x\to 3^-} m(x) = 2^3 = 8$, and the right-hand limit uses $x+5$ so $\lim_{x\to 3^+} m(x) = 3+5 = 8$. Since both one-sided limits exist and are equal, $\lim_{x\to 3} m(x) = 8$.

Finally compare the limit to the function value: $\lim_{x\to 3} m(x) = 8$ and $m(3) = 8$, so these agree. Since all three conditions of continuity hold ($m(3)$ is defined, the limit exists, and they are equal), $m$ is continuous at $x=3$, and no discontinuity classification applies.