Is Ohm's law valid for all frequencies? From continuity equation and Ohm's law it it's possible to say that: 
$$ \nabla \cdot (\sigma \mathbf{E}) = -\frac{\partial \rho}{\partial t} \Rightarrow -\frac{\partial \rho}{\partial t} \frac{1}{\sigma} = \nabla \cdot \mathbf{E}$$
and using Gauss's law:
$$ -\frac{\partial \rho}{\partial t} \frac{1}{\sigma} = \frac{\rho}{\epsilon_0}$$
This is a first-order ordinary differential equation, so its solution is given by:
$$ \rho (\mathbf{r},t) = \rho(\mathbf{r},0) e^{-\frac{\sigma t}{\epsilon_0}} $$
From that expression we can say that $\rho \rightarrow 0$ as $ t $ increases, but also will its derivative, so going back to the continuity equation:
$$  \nabla \cdot \mathbf{J}= -\frac{\partial \rho}{\partial t} \rightarrow 0 $$
For good conductors, this seems to mean that in practically no time 
$$ \nabla \cdot \mathbf{J} \approx 0 $$
This is one of the assumptions for the quasistatic approximation, but as far as I know, this is not always true, for instance while working at very high frequencies. But all the steps made in the derivation were independent from the frequency of $ E $, so why this doesn't apply for all frequencies?
 My intuition dictates me that the problem would be Ohm's law, that may not be valid for very high frequencies as the conductivity starts to behave differently (imaginary part), nevertheless, I've been told that for almost every practical application the conductivity matches its DC value, so Ohm's law can be applied, isn't this true? and if it is, when is not possible to say that $ \nabla \cdot \mathbf{J} \approx 0 $ ?
 A: Yes, Ohm's law is valid at finite frequency
$$
 j(\omega)=\sigma(\omega)E(\omega).
$$
This is simply a linear response relation, so the only assumption is that of a sufficiently weak field. A typical approximation for $\sigma(\omega)$ is the Drude form
$$
\sigma(\omega) = \frac{\sigma_0}{1+i\omega\tau}
$$
which is not exact, but valid in kinetic theory. Indeed, in many typical applications you can neglect the frequency dependence, because the time scale
is set by a microscopic collision time $\tau\sim\tau_{coll}$. Note that in real time, the frequency dependent response becomes a convolution integral
$$
j(t) = \int dt'\, G(t-t')E(t').
$$
where $G$ is the retarded (or response) function, the one-sided Fourier transform of $\sigma(\omega)$.
With regard to your manipulations: I think that $\nabla \cdot j=0$ is indeed a good approximation, and that the main correction arises from non-localities in the response that are important for high frequencies and short wavelength (compared to collision frequencies and mean free paths). A typical problem you find discussed in texts book on EM is the skin effect. Indeed, the skin depth $\delta=\sqrt{2/(\mu\omega\sigma)}$ is governed (approximately) by the DC conductivity $\sigma$ ($\mu$ is the permeability). 
A: If you assume that the only property of the material is that of conductivity, then yes. Ohms law is valid at all frequencies.
But we know real materials have properties of capacitance and inductance as well as conductance and which also depend on the geometry as well as the material itself. So in this case, no, you need to include additional physical models that address these properties. 
In general impedance, which encompasses capacitance, inductance as well as conductance will change according to frequency.
