Inductance in superconducting wires (and the math) I'm very curious about inductors in general, and how similar they are to capacitors.

In the above picture the formula of current over time in an inductor with wire resistance $R$ is written.
I'm curious whether this formula works for superconductors.
As I've been taught, superconductors have a resistance of exactly 0, and the formula is not mathematically defined at 0, so how do we define this state?
My professor told me that I have to calculate the limit approaching 0, but I found that doesn't make sense when resistance is exactly 0; So if someone could explain to me I would be thankful.
Edit: Thanks for all the answers they were all helpful in some way or another, but i chose the one i felt was most helpful.
 A: Let us try as your professor suggested as lim R->0 the equation becomes 0/0 which is undetrmied so let's use L'Hospital's Rule. we get
$$I(t)=\frac{V_b }L .t e^{-tR/L}=\frac{V_b.t}L \, .$$
Yes because an ideal inductor is short circuit for steady state.   as you might know inductive reactance $X_L=j \omega L$ as it's DC, frequency is zero it's practically a short circuit since it violates Kirchhoff's voltage law hence a large current flows in the circuit.
A: The case you presented in within the "regular" ohmic regime, for which the Ohm's law has the following formulation -
$$J=\sigma E$$
Here $J$ is current density, $\sigma$ is the conductivity (inverse of resistivity) and $E$ is the electric field (see Drude model for the intuitive derivation of it). Integration of this relation recreates the more popular version, $I=V/R$. Notice that this is a simplified relation, which doesn't hold in general but in certain parameters regime. Now, the meaning of this relation is interesting - the current is proportional to the force! The analogy from mechanics is that of terminal velocity of a mass subjected to a constant force. This is of course great deviation from Newton't law, where the acceleration is proportional to the force, and not the velocity.
A great advancement in understanding superconductivity was to recreate "Newton's law for currents" (which is one of the London equations), there the electric field (~the force) causing the current density to increase (~acceleration),
$$\frac{\partial J}{\partial t}=\sigma E$$
Practically speaking, any physical circuit always has resistive parts (the battery, the wires at least), and even the measurement devices (ampermeter) are not ideal. Thus the finite resistance of the circuit will balance the voltage drop of the battery, and you will see similar properties.
Notice that superconductivity is a quantum phenomena, and classical explanations are at best good intuitive phenomenological models.
A: Let's start from the initial equation
$$V_b = IR + L \frac{dI}{dt} \, .$$
When the resistance is zero, the equation reduces to
$$V_b = L \frac{dI}{dt} \, .$$
Integrating, we get $I = V_b t / L$ (assuming initial current was zero). There was no need to take any limits of $0/0$ form.
To keep the voltage constant, the current must change in such a way that the rate of change of current is constant, i.e. current is a linear function.
A: If you look at what happens to maximum (final) current as $R$ goes to zero, it increases without any limit. So if you connect superconducting coil to ideal source of voltage, current will increase until it is so high that it destroys the superconducting state in the coil. Then the coil resistance will massively increase in a short time and energy stored in the superconductor will turn to heat, possibly melt the coil. Superconductors can't carry too high currents.
