The short answer: No, classical mechanics is not recovered in the $\hbar \rightarrow 0$ limit of quantum mechanics.
The paper What is the limit $\hbar \rightarrow 0$ of quantum theory? (Accepted for publication in the American Journal of Physics) found that
Our final result is then that NM cannot be obtained from QT, at least by means of a mathematical
limiting process $\hbar \rightarrow 0$
[...]
we have mathematically shown that Eq. (2) does not follow from Eq. (1).
"NM" means Newtonian Mechanics and "QT" quantum theory. Their "Eq. (1)" is Schrödinger equation and "Eq. (2)" are Hamilton equations. Page 9 of this more recent article (by myself) precisely deals with the question of why no wavefunction in the Hilbert space can give a classical delta function probability.
First, the classical and semiclassical adjectives are not quite synonyma. "Semiclassical" means a treatment of a quantum system whose part is described classically, and another part quantum mechanically. Fields may be classical, particle positions' inside the fields quantum mechanical; metric field may be classical and other matter fields are quantum mechanical, and so on.
Also, we often treat the "quantum part" of the semiclassical treatment in another approximation – where we take the leading classical behavior plus the first quantum correction only. For this part of the system, the "semiclassical" therefore means "one-loop approximation" (like in the WKB approximation).
Now, the laws of quantum mechanics may be shown to imply the laws of classical physics for all the "classical questions" whenever $\hbar\to 0$. More properly, $\hbar\to 0$ indeed means $J / \hbar \to \infty$ for all the normal "angular momenta" $J$, actions $S$ (instead of $J$), and everything else with the same units. So yes, indeed, the $\hbar\to 0$ classical limit and the limit of the large quantum numbers is the same thing. It is not kosher to ask whether a dimensionful quantity such as $\hbar$ is much smaller than one; whether the numerical value is small depends on the units. So we must make these claims about "very small" or "very large" dimensionless, and that's why we need not just $\hbar$ but also $J$ or $S$ of the actual problem, and that's why all the inequalities dictating the classical limit that you mentioned are equivalent.
In this limit, the spectra become so dense that the observables (such as the energy of the hydrogen atom) are effectively continuous even though they are discrete in the exact quantum treatment. The Heisenberg equations of motion for the operators reduce to the classical equations of motion. Decoherence guarantees that with some environment, the diagonal entries of the density matrix may be interpreted as classical probabilities, and the off-diagonal ones quickly go to zero. We may always imagine that the wave functions in this limit are "narrow packets" whose width is negligible and whose center moves according to the classical equations. It just works.
One should understand all aspects of this proof that "classical physics is a limit of quantum mechanics", what it assumes, how we should ask the questions and translate them from one formalism to another, and so on. But at the end, the fact that this statement holds is more important than some technical details of the proof.
Historically, the Hamilton-Jacobi equation is a way to describe classical physics because it was discovered and shown equivalent to classical physics long before the quantum theory was first encountered. Mathematically, you can see that the Hamilton-Jacobi equation only contains the quantities we may actually measure with classical apparatuses such as $S,t,V,m$ etc. and it doesn't depend on $\hbar$ at all – even if you use the SI units, for example – which proves that the equation is independent of quantum mechanics.
There are lots of things to say about the classical limit of quantum mechanics and some more specific classes of quantum mechanical theories, see e.g.
http://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html?m=1
Best Answer
Classical limit may mean several things in quantum mechanics:
Example
As a simple way to get some intuition, one could consider a Gaussian wave packet $$ \psi(x)=\frac{1}{(2\pi\sigma^2)^{1/4}}e^{-\frac{(x-x_0)^2}{4\sigma^2}}, |\psi(x)|^2=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-x_0)^2}{2\sigma^2}}, \langle x\rangle = \int dx x|\psi(x)|^2 = x_0. $$ Then $$ \langle V(x)\rangle = V(x_0)\langle e^{\log V(x)-\log V(x_0)}\rangle \approx V(x_0)\langle e^{\alpha(x-x_0)+\frac{1}{2}\beta (x-x_0)^2}\rangle = V(x_0)\frac{1}{\sqrt{1+\sigma^2\beta}}e^{\frac{\sigma^2\alpha^2}{1+\sigma^2\beta}},\\ \alpha = \frac{d}{dx}\log V(x)|_{x=x_0},\beta = \frac{d^2}{dx^2}\log V(x)|_{x=x_0}. $$ The center-of-mass description then holds for $$ \sigma^2|\beta| = \sigma^2\left|\frac{d^2}{dx^2}\log V(x)|_{x=x_0}\right|\ll 1,\\ \sigma^2\alpha^2 = \sigma^2\left|\frac{d}{dx}\log V(x)|_{x=x_0}\right|^2\ll 1, $$ i.e., when the potential is smooth on the characteristic scale of the wave function change.