e^-it as t approaches infinity

Question

How do I evaluate $$ \lim_{t\rightarrow \infty } e^{-it}$$

I feel like it should be 0, but I'm not sure if the $i$ changes things.

Assuming that $t$ is real, the function $t \mapsto \mathrm{e}^{-it}$ is periodic, and does not have a limit at infinity. — Xander Henderson, Oct 03 '18 at 04:31
Complex numbers can be represented in the form $re^{i\theta}$ where $r$ is a radius from the origin and $0\leq\theta<2\pi$ is the angle of rotation from the positive real axis. But this function is periodic in $\theta$; you're just traveling around the boundary of a circle as you increase $\theta$. — kcborys, Oct 03 '18 at 04:37

score 3 · Answer 1 · answered Oct 03 '18 at 05:12

3

Hint:

$$e^{iat}=\cos at + i \sin at$$

answered Oct 03 '18 at 05:12

score 2 · Answer 2 · answered Oct 03 '18 at 05:33

2

$e^{it}=(-1)^{n}$ when $t=n\pi$ so the limit does not exist.

answered Oct 03 '18 at 05:33

Kavi Rama Murthy

score 1 · Answer 3 · answered Oct 03 '18 at 05:18

1

Here is how you can interpret this limit using argand plane. The solution region is a square and there is no definite value for this limit

answered Oct 03 '18 at 05:18

Math Maniac

3 Answers3