They're two different limits in which two different constants are sent to zero and the resulting limiting theory has different names.
However, both of them are limits for dimensionful constants and the analogy is perfect.
The properly derived $\hbar\to 0$ limit of a quantum mechanical theory is a classical theory – its classical limit – in the very same sense in which the properly derived $k\to 0$ limit of the laws of statistical mechanics produce the laws of thermodynamics.
Now, the limit $k\to 0$ and $N\to \infty$ really means the same thing because the implicit assumption in all these limiting procedures is that the macroscopic quantities known from the everyday life are kept finite – and essentially fixed. This is especially the case for the energy and temperature. The thermal energy of $N$ atoms is something like
$$ E=3kTN/2 $$
If both $E,T$ are fixed, it's clear that the $k\to 0$ limit is exactly the same thing as the $N\to\infty$ limit. The thermodynamic limit simply means that we're neglecting all effects in a fixed-energy system that are caused by the finiteness of the number of atoms – or, equivalently, the finiteness (nonzero value) of the contribution of a single atom to the whole (which is proportional to $k$).
One has the freedom to describe the limit in many ways in the quantum mechanical case, too. We may say that the classical limit appears as $\hbar\to 0$. But we may also say that the classical limit emerges when $N_J\to \infty$ where $N_J=J_z/\hbar$, for example. When the angular momentum (or the action $S$) is written as a multiple of $\hbar$, the dimensionless coefficients' condition $N_J\to\infty$ is the same thing as $\hbar\to 0$ because their product is fixed. Classical physics needs to neglect all effects caused by the situation when the overall quantities' describing the system are so small that they're comparable to $\hbar$ (which is the regime where the quantum phenomena become important). Again, we're comparing two things, so saying that one of them is infinitely larger than the other is the same thing as saying that the other one is infinitely smaller.
In both situations, one has many options what $N$ or $N_J$ may exactly be. But in both cases, the limit for the dimensionful constant going to zero, whether it's $k$ or $\hbar$, is equivalent to some dimensionless numbers' (those that measure how much the system is larger relatively to the statistical mechanical or quantum "basic blocks" where the more general theory shows in its full glory) going to infinity.
Because the ratio of probabilities of an entropy-changing process and its time reversal goes like $\exp((S_B-S_A)/k)$, we see that for fixed $S_A,S_B$ in macroscopic units, the ratio becomes strictly infinite. So in thermodynamics, i.e. the thermodynamic limit of the statistical considerations, a decreasing entropy is strictly impossible.
Finally, let me mention that the nonrelativistic limit is analogous to the two limits above, too. We may say that the limit involves $c\to\infty$ which is clearly the same thing as $1/c\to 0$: it plays the same role as $k\to 0$, for example. However, we may also say that the actual velocities are much smaller than $c$ in the limit, so $\beta=v/c\to 0$ or $1/\beta\to \infty$. That's analogous to $N\to\infty$ or $N_J\to\infty$ above.