how does current go to infinity in an ideal LC circuit at resonance?
I understand the impedance goes to zero,which explains the steady
  state current as infinite. But can you explain the same by simple
  circuit analysis?I mean the voltage across inductor is Vsinωt - Vc = L
  di/dt, so based on derivative being positive, the current
  increases.But what makes the current go to infinite?


Yes, the net impedance falls to zero in a series resonant circuit when R is zero and, the voltage across each of L and C also rises to infinity - it couldn't be anything else or you would not get infinite current. That's the simple circuit analysis - trying to equate \$V_L\$ to the input voltage minus \$V_C\$ doesn't really help other than remind you that the two voltages (across L and C) are oppositely phased i.e. they are 180 degrees apart.

Should you think of the driving source as a constant current (just to prevent infinities and keep things sensible) that current appears to conduct thru a dead-short-circuit at resonance and that current will produce an equal finite voltage magnitude across each element determined by the components impedance (equal at resonance).

The current will reach infinity non-instantaneously because resonance is at a pure sinewave and a pure sinewave has to have been in existence a very long time ago!

Energy transfer is exactly the same as when not exactly at resonance or with a little bit of resistance except the numbers are infinity (not helpful to analyse in my book).

There are initial conditions - the application of a sinewave via a switch is an initial condition but, if you mean a dc voltage across the capacitor then this will continue to be present (on average) because the energy contained in it has nowhere to be dissipated (neither the perfect voltage source nor inductor) can do this. In a real circuit this DC "energy" will be burnt in the non-zero resistance of the wires.