Evaluate the following limit in $\mathbb{D}'(\mathbb{R}^n)$: $$\lim_{\varepsilon \to 0^+} \, \frac{e^{\frac{i |x|^2}{4 \varepsilon}}}{(4 \pi \varepsilon)^{\frac{n}{2}}}$$ where $|x|$ is the Euclidean norm of $x$ in $\mathbb{R}^n$.
I was dealing with this limit because it was left as an exercise from the book, but i have no clue on how to solve it. As an hint, the book suggests to start with $n=1$, so i tried to solve it using the lebesgue's dominated convergence theorem but i can't still find the correct answer.
(for $n=1$) let $$F_\varepsilon(x)=\int_0^{\varepsilon^{-1/2} x} \frac{e^{i y^2/4}}{(4\pi)^{1/2}}dy$$ The main point is that $$\lim_{x\to \infty} F_\varepsilon(x) =sign(x) C $$ converges (the integral is some kind of alternated series). Whence $$\lim_{\epsilon\to 0}F_\varepsilon=sign(x) C$$ in the sense of distributions, from which $$\lim_{\varepsilon \to 0^+} \, \frac{e^{i x^2/(4 \varepsilon)}}{(4 \pi \varepsilon)^{1/2}}=\lim_{\epsilon\to 0}F_\varepsilon'=(sign(x) C)'=2\delta(x) C$$ in the sense of distributions.
To find $C$: $\int_0^\infty e^{-x^2}dx = \frac12\sqrt{\pi}$ gives $\int_0^\infty e^{-a^2 x^2}dx = \frac1{2a}\sqrt{\pi}$. By analytic continuation it stays true for $a\in \Bbb{C}^*,\Re(a)\ge 0$ so that $C=\frac12 e^{i\pi/4}$.