Does this limit make any sense? Does it make sense to take the following limit?
$$\lim_{\phi\to\infty}e^{i \phi}=?$$
And if yes, what does it yield?
EDIT:
I vaguely remember someone mentioning that this limit gives zero in a distributional sense, since all the rotations 'balance' to zero. Is this point of view any good?
 A: Yes, such limits are well-defined. However, the limit in question does not exist due to oscillation. This is merely a complex notation for $\lim_{\phi \rightarrow \infty} \cos(\phi) + i \lim_{\phi \rightarrow \infty} \sin(\phi) $ which clearly does not converge.
A: The limit makes sense is the sense that for each real number $\phi$ you can evaluate $e^{i\phi}$.
But the limit does not exist because there does not exist a complex number $L$ such that for all $\epsilon > 0$ there is a $N$ such that whenever $\phi \geq N$ then $\lvert e^{i\phi} - L\rvert < \epsilon$. 
Note for example that with $\epsilon = 1$ and with given $N$ you can find an integer $m$ such that $2\pi m \geq N$ (and then also $2\pi m + \pi \geq N$). But $e^{2\pi m} = 1$ and $e^{2\pi m + \pi} = -1$... 
A: The limit does not exist. As $\phi\rightarrow\infty$, the value $e^{i\phi}$ "turns and turns and turns and turns" over the complex unit circle without converging to any particular complex number.
A: As for your "distributional sense", it is true that $$\lim_{\phi\to\infty}\frac 1 \phi \int_0^\phi e^{i\theta}\,d\theta=0,$$ but I'm not sure why this is interesting.
