Yes, it does follow that f'(0) exists.

Hmm, you never said whether this has to be proven or just shown, so my explanation will not be all that rigorous.

Basically, as f is continuous on all the reals, its derivative cannot have any

**removable** discontinuities, i.e. discontinuities where the lefthand and righthand limits agree as x->c but f(c) does not exist. This is because all removable discontinuities are removed in the limit.

To clarify this point, consider

If the lefthand limit agrees with the righthand limit, and f and g are both continuous, then the limit is defined as a number.

Making use of this point, for f'(0) not to exist, there must be a removable discontinuity at x = 0, yet we know there cannot be, since all removable discontinuities are removed in the process of differentiation, i.e.

Thus f'(0) exists.