Capacitor acts like short circuit at t=0, the reason that capacitor have leading current in it. The inductor acts like an open circuit initially so the voltage leads in the inductor as voltage appears instantly across open terminals of inductor at t=0 and hence leads.
A capacitor looks like an open circuit to a steady voltage but like a closed (or short) circuit to a change in voltage. And inductor looks like a closed circuit to a steady current, but like an open circuit to a change in current. You probably should put this as an answer, as I believe that is what the OP is looking for.
(A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.
V(t) = V(−Rt/L)e V (t) = At t = ∞ t = ∞, V = 0 V = 0 so the inductor behaves as an short circuit. Because capacitors store energy in the form of an electric field, they tend to act like small secondary-cell batteries, being able to store and release electrical energy.
Short Answer: Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor) Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor) Long Answer:
The inductor acts like an open circuit initially so the voltage leads in the inductor as voltage appears instantly across open terminals of inductor at t=0 and hence leads. simply remember capacitor rises voltage from 0 to high,so at intitally at ov capacitor acts as short ckt and for high voltage cap acts as open ckt, reverse in case of inductor