asymptotic Definition and 1 Threads

When one does the phase transformation ##\psi(\vec{x},t)=R(\vec{x},t)\exp^{iS(\vec{x},t)/\bar{h}}## For this transformation to be valid doesn't it need to have the same asymptotic behaviour ##x \to \pm \infty##, and also at ##x=0 ## as the original wave-function ##\psi## does? How come this is...