# What is the limit of e^-ix

What is the limit of $e^{-ix}$ as x tends to infinity?

Mark44
Mentor
What is the limit of $e^{-ix}$ as x tends to infinity?
What does ##e^{-ix}## represent? IOW, for a given x value, what does ##e^{-ix}## evaluate to?

$cos(x) - isin(x)$

Ray Vickson
Homework Helper
Dearly Missed
$cos(x) - isin(x)$

OK, so what do YOU think the limit should be, if anything?

Well this is exactly my problem, I dont know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I dont know if this is now the right place to ask this but if anyone can help that would be great.

mathman
I can't comment on the physics question. However the original math question is answerable - there is no limit. You can envision it geometrically as being points on the unit circle in the complex plane. As x becomes infinite the point simply goes around the circle indefinitely.

Ray Vickson
Homework Helper
Dearly Missed
Well this is exactly my problem, I dont know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I dont know if this is now the right place to ask this but if anyone can help that would be great.

Do you know what it means when we say that a function, f(x), has a limit as x → ∞? Never mind the "epsilon-delta" stuff; just give an intuitive description.

Alternatively, think of the graph y = cos(x). Do the y-values settle down to a fixed value as x becomes larger and larger?

Matterwave
Gold Member
Well this is exactly my problem, I dont know. Perhaps I should have mentioned that I have considered the limit in terms of cos and sin and I'm not just asking you out of laziness. I would be inclined to say the limit does not exist. The reason I need to know is I am answering a question on square well potentials where solving schrodingers equation yields $\psi(x)=Ae^{ikx} +Be^{-ikx}$ outside of the well which in the region to the leftof the well simplifies to $Ae^{ikx}$ and I was wondering if this is because the wave function equals zero as x tends to minus infinity which implies B=0. I dont know if this is now the right place to ask this but if anyone can help that would be great.

A (infinite) square well potential should not have waves outside of the well. Outside the well, the wave function should just be 0. A finite square well can have a non 0 wave function outside the well, but they should exponentially decay instead of oscillate (assuming a bound state). Recheck your answers.

If you are dealing with scattering states, then the wave function must be a wave-packet, not a plane wave since plane waves are not normalizable.

Mark44
Mentor
What does ##e^{-ix}## represent? IOW, for a given x value, what does ##e^{-ix}## evaluate to?

$cos(x) - isin(x)$
No, I was looking for a more specific answer, which @mathman gave you in post #6. In my question I specified "for a specific x value," so your response should have taken that into account.