There is a way in which the Schrödinger equation on the limit $\hbar \to 0$ reduces to the Hamilton-Jacobi equation. In that limit, it turns out that it is not so much the wave function itself that satisfies the Hamilton-Jacobi equation but its phase. If a collision state of the form is considered:
$$\psi(\boldsymbol{x},t) \approx e^{iS(\boldsymbol{x},t)/\hbar}$$
Where $S(\boldsymbol{x},t)$ is the action of the particle, substituting this equation into the Schrödinger equation:
$$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\left( \frac{\partial^2 \psi}{\partial x^2}+\dots+\frac{\partial^2 \psi}{\partial z^2}\right) + V(\boldsymbol{x})\psi$$
results in the following equation for the action:
$$-\frac{\partial S}{\partial t} e^{iS/\hbar} = \frac{\hbar^2}{2m}\left[ \left( \frac{\partial S}{\partial x}\right)^2 +\dots + \left( \frac{\partial S}{\partial z}\right)^2\right]e^{iS/\hbar} + V(\boldsymbol{x})e^{iS/\hbar} - \frac{i\hbar}{2m}(\Delta S) e^{iS/\hbar}$$
By canceling the exponential factor and taking the real part of this equation, we have:
$$\frac{\partial S}{\partial t} + \frac{\hbar^2}{2m}\left[ \left( \frac{\partial S}{\partial x}\right)^2 +\dots + \left( \frac{\partial S}{\partial z}\right)^2\right] + V(\boldsymbol{x}) = 0$$
Which is the classical Hamilton-Jacobi equation for a particle in a potential. The imaginary part null, in the limit $\hbar \to 0$:
$$\frac{i\hbar}{2m} \Delta S = 0$$
If we start from the totally general expression:
$$\psi(\boldsymbol{x},t) = A(\boldsymbol{x},t)e^{iS(\boldsymbol{x},t)/\hbar}$$
in the limit $\hbar \to 0$, the additional terms cancel out and we simply arrive also at the Hamilton-Jacobi equation for the complex phase of the wave function.