smc

Note that $h$ is a length, while $m$ is a dimensionless constant. Thus, $h$ and $m$ cannot be added, and $B$ and $E$ are not proper answers, since they both contain $h+m$. Thus, we only concern ourselves with answers $A,C,D$. If $m$ is a very, very large number, then Moonbeam will have to run just over $h$ meters to reach Sunny. Or, in the language of limits: ${\lim }_{m\to \infty }d(m)=h$, where $d(m)$ is the distance Moonbeam needs to catch Sunny at the given rate ratio of $m$. In option $A$, when $m$ gets large, the distance gets large. Thus, $A$ is not a valid answer. In option $C$, when $m$ gets large, the distance approaches $0$, not $h$ as desired. This is not a valid answer. (In fact, this is the distance Sunny runs, which does approach $0$ as Moonbeam gets faster and faster.) In option $D$, when $m$ gets large, the ratio $\frac{m}{m-1}$ gets very close to, but remains just a tiny bit over, the number $1$. Thus, when you multiply it by $h$, the ratio in option $D$ gets very close to, but remains just a tiny bit over, $h$. Thus, the best option out of all the choices is $\boxed{\frac{hm}{m-1}}$.