This question comes from Fourier Transforms, specifically the evaluation of $\mathcal{F}(e^{2\pi iat})$. Normally I see this derived by first finding the Inverse FT of a delta function, i.e. \begin{align} \mathcal{F}^{-1}(\delta(\omega-a)) &= \int_{-\infty}^\infty \delta(\omega-a) e^{2\pi i \omega t}\,d\omega = e^{2\pi i at} \end{align} But suppose we wanted to be stubborn and directly evaluate \begin{align} \mathcal{F}(e^{2\pi i a t}) &= \int_{-\infty}^\infty e^{2\pi i a t} e^{-2\pi i \omega t} \,dt \end{align} It should work, right? Why wouldn't it? If we continue, we get to
\begin{align} \mathcal{F}(e^{2\pi i at}) &= \lim_{x \to \infty}\int_{-x}^x e^{2\pi i (a - \omega) t} \,dt \\ &= \frac{1}{\pi(a-\omega)}\lim_{x \to \infty}\sin(2\pi(a-\omega) x) \\ &= \frac{1}{\pi (a - \omega)}\lim_{x \to \infty} \sin(x) \end{align}
Now, how do we convincingly evaluate $\displaystyle \lim_{x \to \infty} \sin(x)$? If I didn't have the context of evaluating a Fourier Transform, I would simply answer that the limit doesn't exist.
However, in this case I know what the answer should be, but I don't at all understand why it works or how to make it rigorous. It seems to contradict my previous understanding of limits.