An example of distributional limit I have to evaluate the following limit in $\mathcal D'(\mathbb{R}^n)$, with $n\in\mathbb{N}$.
$$
\lim_{k\to \infty}k^{n/2} \exp(ik|x|^2)
$$
I gave it a try, but it seems "too easy"
$$
\forall\,f\in\mathcal D(\mathbb{R}), \quad \forall\,k>0 \qquad |(e^{ik|x|^2},f)|=\left|\int_{\mathbb{R}^n}\mathrm{d}^nxe^{ik|x|^2}f\right|\leq\|f\|_{L^1}
$$
so $\displaystyle \lim_{k\to\infty}|k^{n/2}e^{ik|x|^2}|\leq \lim_{k\to\infty}k^{n/2}\|f\|_{L^1}=\infty$
 A: 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.)
