An example of distributional limit

Just following up on @reuns' observation, and responding to some comments by the questioner:

Yes, Fourier transform applies best (but not only!) to tempered distributions, in the sense that it maps tempered distributions to tempered distributions. Yes, (integration-against) $e^{ik|x|^2}$ is a tempered distribution: a standard inequality gives $$ \Big|\int_{\mathbb R^n} e^{ik|x|^2}\cdot f(x)\;dx\Big| \;=\; \Big|\int_{\mathbb R^n} {e^{ik|x|^2}\over (1+|x|^2)^{n/2+\epsilon}}\cdot (1+|x|^2)^{n/2+\epsilon}f(x)\;dx\Big| $$ $$ \;\le\; \int_{\mathbb R^n}{dx\over (1+|x|^2)^{n/2+\epsilon}} \cdot \sup_x (1+|x|^2)^{n/2+\epsilon}\cdot |f(x)| $$ The latter sup is one of the Schwartz-space seminorms, so this proves that that integration-against functional is a tempered distribution (etc.) The style of this computation is a thing that should be made familiar...

The computation of the Fourier transform (as aptly suggested by @reuns) is not trivial, since a rigorous version of it requires a bit of fancier stuff, (also worth learning: holomorphic vector-valued functions as such), but/and is essentially, up to various irrelevant constants, (integration-against) $e^{-i|x|^2/k}$. Relatively elementary inequalities show that this goes to (integration-against) $1$.

Thus, Fourier inversion (for tempered distributions) gives that the original limit (because Fourier transform is continuous on tempered distributions, it preserves limits!) is a constant multiple of $\delta$. !!!! :)

The constant can be determined by applying the sequence of distributions to something like $e^{-\pi |x|^2}$... :)

(I think it's not so easy to see this outcome rigorously without using Fourier transform.)