I am trying to prove that $$ \lim_{y\rightarrow 0} \delta(y-x) = \delta(x) .$$
To justify this problem, this comes from the orthogonality of the position eigenstates in quantum mechanics. Indeed, we have that $\langle x|y \rangle = \delta(x-y)$, but what happens if $y$ approaches $x$?
This looks trivial, but what if $x=0$? If we think of the delta as something which is equal to zero everywhere but in zero, then we should have $\lim_{y\rightarrow 0} \delta(y) = 0$, while from the identity above we would have $\lim_{y\rightarrow 0} \delta(y) = \delta(0)$. I think that, in this case, the problem is in the definition of the delta.
So I tried the following:
consider a continuous function $f$: $$ \int_{-\infty}^{+\infty} f(x)\delta(x)dx = f(0) = \lim_{y\rightarrow 0} f(y) = \lim_{y \rightarrow 0} \int_{-\infty}^{+\infty} f(x)\delta(x-y)dx = \int_{-\infty}^{+\infty} f(x) \left(\lim_{y\rightarrow 0}\delta(x-y) \right) dx $$ and since it has to hold for every continuous $f$, we have $$\lim_{y\rightarrow 0}\delta(x-y) = \delta(x) .$$ However, I'm not really sure that everything works since I'm swapping the limit and the integral, and this could require more assumptions.
I tried to use also other definitions of $\delta$ as distribution but I always have to swap limits and I'm not sure I can have enough assumptions to do it.
To conclude, I would like to have an answer to $\lim_{y \rightarrow 0} \delta(y) = ?$, but since it could be not well defined, feel free to use related topics like the orthonormality of states or other definitions of delta.
The Dirac delta at a point $y\in\Bbb R^d$, $\delta_y(x) := \delta(x-y)$ is defined as a distribution by its action on test functions $\varphi\in C^\infty_c$ through the formula $$ \langle \delta_y,\varphi\rangle = \int_{\Bbb R^d} \varphi(x)\,\delta_y(\mathrm d x) = \varphi(y). $$ In particular, since $\varphi$ is continuous, $\varphi(y)\to\varphi(0)$ when $y\to 0$, which implies that $$ \langle \delta_y,\varphi\rangle \underset{y\to 0}{\to} \langle \delta_0,\varphi\rangle $$ that is $$ \delta_y \underset{y\to 0}{\to} \delta_0 \text{ in the sense of distributions,} $$ which you can write symbolically, $\delta(x-y) \underset{y\to 0}{\to} \delta(x)$.