This doesn't directly answer your question, but you may be interested to know that if you carry on the limit to infinite $n$, then the limit is zero. In fact, the ratio above is zero for every infinite $n$. The reason is that most infinite categories are disconnected, a fact that I personally find surprising, but there it is:

> $\#\mathsf{Cat}_{\mathrm{conn}}(\kappa) < 2^{\#\mathsf{Cat}_{\mathrm{conn}}(\kappa)}\leq \#\mathsf{Cat}(\kappa)$

where 

 - $\kappa$ is any infinite cardinal,
 - $\#\mathsf{Cat}(\kappa)$ is the number, up to equivalence, of categories with _fewer_ than $\kappa$ morphisms,
 - $\#\mathsf{Cat}_{\mathrm{conn}}(\kappa)$ is the number, up to equivalence, of _connected_ categories with _fewer_ than $\kappa$ morphisms.

Here a category is _connected_ if there is a zigzag of morphisms between any two objects. Note that monoids are connected. On a technical note, notice that I'm using a strict inequality in my definitions where you used a non-strict inequality; I do this because strict inequalities are more general in the infinite case. (To recover the version with non-strict inequalities, just insert $\kappa^+$ wherever I have $\kappa$).

The argument is very simple. Two categories are equivalent if and only if there is a bijection of their connected components pairing equivalent components. So if $\{\mathcal{C}_x\}_{x \in X}$ is a set of pairwise inequivalent connected categories, then $\{\coprod_{x \in S} \mathcal{C}_x\}_{S\subseteq X}$ is a set of pairwise inequivalent categories.

Actually, it doesn't take much more work to get precise numbers. On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures. On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. Hence we have
> $\#\mathsf{Cat}_{\mathrm{conn}}(\kappa) \geq \kappa \qquad \& \qquad \#\mathsf{Cat}(\kappa) = 2^\kappa$

If the power operation $\kappa \mapsto 2^\kappa$ is injective, then that "$\geq$" can be improved to "$=$", but I'm not sure if it can in general.