smc

This problem is really finding the range of a function with a restricted domain. Dividing $x+2$ into $2x+3$ yields $2-\frac{1}{x+2}$. Since $x\ge 0$, as $x$ gets larger, $\frac{1}{x+2}$ approaches $0$, so $\frac{2x+3}{x+2}$ approaches $2$ as $x$ gets larger. That means $M=2$. Since $\frac{1}{x+2}$ can never be $0$, $\frac{2x+3}{x+2}$ can never be $2$, so $M$ is not in the set $S$. For the smallest value, plug in $0$ in $\frac{2x+3}{x+2}$ to get $\frac{3}{2}$, so $m=\frac{3}{2}$. Since plugging in $0$ results in $m$, $m$ is in the set $S$. Thus, the answer is $\boxed{\text{m is in S, but M is not in S}}$.