Limit $\lim_{x\to\infty}e^{i a x}$? Is there anything meaningful one can say about the limit
$$\lim_{x\to\infty}e^{i a x}$$
for $a\in\mathbb C$ ? Thanks for any suggestion!
 A: If we let $a = \alpha + \beta i$, then
\begin{align*}
e^{iax} &= e^{i(\alpha + \beta i)x} \\
&= e^{\alpha xi - \beta x} \\
&= e^{-\beta x} e^{\alpha x i}
\end{align*}
where the first factor $e^{-\beta x}$ is the modulus and scales the second factor $e^{\alpha x i}$ (the argument) which denotes a complex number on the unit circle that revolves around the origin as $x$ is increased with the rotational speed and direction determined by the value of $\alpha$ (positive means it revolves counter-clockwise, negative means it revolves clockwise, zero means it stays fixed at the real number $1$).
Hence if $\beta > 0$, we have the modulus decaying to zero as $x$ increases, and so $e^{iax}$ spirals toward zero, and we have limit $0$.
If $\beta < 0$, we have a positive coefficient for $x$ in $e^{-\beta x}$ and so the modulus grows without bound and we get $e^{iax}$ spiraling out to infinity.
If $\beta = 0$, we have the modulus fixed at $1$ and so $e^{iax}$ forever orbits the origin at unit distance and hence the limit is undefined (unless $\alpha = 0$ also, in which case $e^{iax}$ is trivially equal to $1$ for all $x$).
A: If we interpret the limit classically, then the limit exists if and only if $\text{Im}(a)<0$.
However, note that for any function $\phi(a)$ that is $L^1$ and with compact support, the Riemann-Lebesgue Lemma guarantees that 
$$\lim_{x\to \infty}\int_{-\infty}^\infty \phi(a)e^{iax}\,da=0$$
Hence, for $a\in \mathbb{R}$, $\lim_{x\to \infty}e^{iax}\sim 0$ as a distribution.
