I have heard that one can recover classical mechanics from quantum mechanics in the limit the $\hbar$ goes to zero. How can this be done? (Ideally, I would love to see something like: as $\hbar$ goes to zero, the position wavefunction reduces to a delta function and that the Schrodinger equation/Feynman path integral reduces to the Newtonian/Lagrangian/Hamiltonian equations of motion.)
[Physics] Classical limit of quantum mechanics
classical-mechanicsquantum mechanics
Related Solutions
The semiclassical limit you're describing says that the amplitude for a particle to get from here to there in a set time is equal to the exponential of the classical action for the corresponding classical trajectory. In symbols this reads $$\langle x_b|U(T)|x_a\rangle=\int \mathcal{D}\varphi e^{iS[\phi]/\hbar} \approx e^{{i}S[\varphi_\textrm{cl}(x_a,x_b,T)]/\hbar}.$$ In a general quantum state, however, particles are not "here" and don't end up "there": they have an initial probability amplitude $\langle x|\psi(0)\rangle$ for being at each position $x$ at time $t=0$ and will have a final probability amplitude $\langle x|\psi(T)\rangle$ for being at position $x$ at time $T$. To apply the approximation, you pull out the propagator and insert a resolution of the identity: $$\langle x|\psi(T)\rangle=\int dy\langle x|U(T)|y\rangle\langle y|\psi(0)\rangle=\int dye^{{i}S[\varphi_\textrm{cl}(y,x,T)]/\hbar}\langle y|\psi(0)\rangle.$$
To get a full semiclassical limit, you also need a semiclassical initial state (since otherwise you've obviously got no hope!). You take, then, a state with (relatively) sharply defined position and momentum (of course, the state will occupy some finite region of phase space but you usually can assume, in these circumstances, that it is small enough), and this will make the amplitudes for points outside the classical trajectory interfere destructively and vanish.
EDIT
So how does this happen? For one, $y$ must be close to the initial position, $y_0$ in order to contribute to the integral. For small displacements of the endpoints, then, the action along the classical trajectory varies as $$\delta S=p_{\varphi,x}\delta x-p_{\varphi,y}\delta y$$ (cf. Lanczos, The Variational Principles of Mechanics, 4th edition, Dover, eqs 53.3 and 68.1, or simply do the standard integration by parts and set $\int\delta L dx=0$ along the classical trajectory). The main contribution of the initial state to the phase is of the form $e^{ip_\textrm{cl}y}$, which means that the integral has more or less the form, up to a phase, $$\langle x|\psi(T)\rangle\approx \int_{y_0-\Delta x/2}^{y_0+\Delta x/2}e^{i(p_{\varphi,y}-p_\textrm{cl})y/\hbar}dy.$$
Here the momentum $p_{\varphi,y}$ is determined by $x$ and (to leading order) $y_0$, since there is a unique classical path that connects them. This momentum must match (to precision $\Delta p\approx \hbar/\Delta x$, which we assume negligible in this semiclassical limit) the classical momentum of the initial state, $p_\textrm{cl}$, and therefore only those $x$'s on the trajectory determined by the initial state will have nonzero amplitudes.
The subtlety is that an arbitrary wavefunction doesn't reduce to a point of the classical phase space in the limit $\hbar \to 0$ (thinking about phase space makes more sense since in the classical limit one should have definite coordinates and momenta).
So one could ask, which wavefunctions do. And the answer is that the classical limit is built on the so-called coherent states -- the states that minimize the uncertainty relation (though I don't know any mathematical theorem proving that it's always true in the general case, but in all known examples it is indeed so). States close to the coherent ones can be thought of as some "quantum fuzz", corresponding to the quasiclassical corrections of higher orders in $\hbar$.
Example of this for the harmonic oscillator can be found in Landau Lifshits.
Regarding the fluid argument. About your remark (1): the $|\psi|^2$ for the stationary state is indeed stationary, but it still satisfies the continuity equation since the current is zero for such states. Your remarks (2) and (3) are quite right because, as I already said, the classical limit can't be sensibly taken for arbitrary states, it is built from coherent states.
And also I must admit that the given fluid argument indeed doesn't provide any classical-limit manifestation. It's just an illustration that "everything behaves reasonably well" to convince readers that everything is OK and to presumably drive their attention away from the hard and subtle point -- it often happens in $\it{physics}$ books, probably unintentionally :). The problem of a nice classical limit description is actually an open one (though often underestimated), leading to rather deep questions, like the systematic way to obtain the symplectic geometry from the classical limit. In my opinion it is also connected to the problem of quantum reduction (known also as the "wave function collapse").
Best Answer
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
"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.