:grumpy: :yuck: Hello there, please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex] S is the action of a particle [tex] = \int L dt[/tex] e.g. when the particle goes to infinity and comes back, kinetic energy blows up, then S blows up.
It does'nt: [tex]e^{ix}=\cos{x}+i\sin{x}[/tex] [tex]\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}[/tex] But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.
I would suspect both your math teacher and physics teacher are right! The map x → e^(ix) is, indeed, discontinuous at +∞, and thus one cannot continuously extend this map to have a value at +∞. (continuous extension is the process that is generally used to justify statements like arctan +∞ = π / 2 or x/x = 1). However... As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.