What is $\epsilon^{i x}$ for $x\in\mathbb{R}$ and $\epsilon\to0^+$?

Question

One could write

$$\epsilon^{i x}=e^{ix\ln(\epsilon)}$$

in which case it's hard to say what should come out. However, it is usually understood that

$$0^{ix}=0$$

meaning "zero to any power is zero". Perhaps in the case of $\epsilon^{i x}=e^{ix\ln(\epsilon)}$ one could argue that the phase rotates around zero a lot in the limit $\epsilon\to0^+$ and therefore averages to zero.

What is the consensus on this? What is $\epsilon^{i x}$ for $x\in\mathbb{R}$ and $\epsilon\to0^+$?

Answer 1

Complex numbers are naughty: they disappoint(or surprise) you always.

Firstly, $y^{ix}$ is multivalued for any $y$, with $0$ a complicated exception. (But since you are now dealing with limit so this exception is not important).

$$c^{ix}=e^{(\ln c+2\pi ik)(ix)}=e^{ix\ln c}\cdot e^{2\pi kx}$$

Recall that $$e^{ix\ln c}=\cos (x\ln c)+i\sin (x\ln c)$$

Unless $x=0$, the limit $c\to0^+$ does not exist due to the oscillating behaviour of $\sin$ and $\cos$.

This is a good example of counter-intuitive results in complex analysis, which is opposite to what we learnt when we were a kid: $0^{\text{anything}}=0$ .

Answer 2

$$\varepsilon^{i x} = \exp(ix \ln \varepsilon) = \cos (x \ln \varepsilon) + i \sin (x \ln \varepsilon)$$

It is clear that it does not converge as $\varepsilon \to 0^+$.

Answer 3

Since $\, |\epsilon^{i x}| = |e^{ix\ln(\epsilon)}| = 1 \,$ the limit as $\, \epsilon \to 0 \,$ does not exist. It just keeps going around the unit circle as $\, \epsilon \,$ diverges from $1$ with $\, x \,$ fixed.