The cutoff-method used for regulating divergences amounts to not integrating over field configurations that have a Fourier-momentum greater than the cutoff in the path integral. However, later on the cut-off is taken to infinity, so in fact we do integrate over all field configurations! The divergences are solved by writing the original bare couplings in terms of experimentally measured couplings, and one consequence is that the experimentally measured couplings change with energy (a change governed by the renormalization group equation) of the experiment. Do the bare couplings change with respect to the cutoff? Or, since everything is going to be taken to infinity anyways, then this is a question that doesn't need to be asked? In the effective-field theory approach, there is a cutoff for the path integral. However, the cutoff is not taken to infinity at the end - this is really a cutoff. As a consequence, loop momenta are only integrated to the cutoff and not to infinity. Therefore there are no divergences in this approach at all. Another consequence is that all bare couplings are finite. These bare couplings change according to the exact same beta function of the physical couplings talked about earlier (when the cutoff is taken to infinity), except instead of the derivative being with respect to the energy of the experiment, it is with respect to the cutoff. What is the relationship between the energy of an experiment and the cutoff used in an effective-field theory approach, such that they obey the same equations? Certainly the coupling constants must have different initial values, because there is a real difference in that one is integrated to infinity, and the other to the cutoff, so the couplings must be different to give back the same answer?