Inductive reactance 0 for dc current? Inductive reactance:
$$X_L=\omega L=2\pi f L$$
is the opposition to the flow of current by an inductor.
Frequency $f$ being $0$ for a DC current, the inductive reactance too is $0$. But doesn't the back EMF produced when a switch is turned on oppose DC current too? Wont that count as an opposition?
 A: When you turn a switch on you are starting the flow of current.  You can Fourrier analyze the current and find many different frequencies, for which there can be inductive reactance.  If the circuit settles to DC, you are correct that there is no inductive reactance.
A: Fundamentally, for a pure inductor $$v_L=L\frac{\mathrm{d}i_L}{\mathrm{d}t},$$
where $v_L$ is the (possibly) time-dependent voltage across the inductor and $i_L$ is the (possibly) time-dependent current through the inductor.
If a sinusoidal current is flowing through the inductor, you will have a sinusoidal voltage across the inductor which is $90^{\circ}$ out of phase with that current if no other impedances are present:
$$i_l=I_{max}\sin(\omega t + \phi_o)\ \to\ v_{L}=\omega LI_{max}\cos(\omega t + \phi_o).$$
When you close an electrical switch to apply a DC voltage across the inductor, the slope of the current through the inductor is huge, even though the current itself must start out at zero.  Same thing when you open a switch: the slope of the current (decreasing) is huge.
As noted by @RossMillikan and commenters, this sudden change in current can be modelled by douing a Fourier decomposition of a square-wave edge step.  You will get many different frequencies, so you will have a sum of reactances which result in a large back-emf.
A: Not necessarily, if within the time constant of the circuit you change the parameters of the circuit as in the winding count you can have on going inductance. as the magnetic flux to current ratio changes as winding's are added or subtracted which according to the Lenz Law will oppose so then you have a current change which falls under inductive reactance.
