# Is the Dirac $\delta$-function necessarily symmetric?

"Delta function" is not a function, but a distribution. Distribution is a prescription for how to assign number to a test function. This distribution may but does not have to have function values in the ordinary sense. In case of delta distribution, it does not have function values.

So statement like

$$ \delta(x) = \delta(-x) \quad\text{for all }x \tag{*} $$ meaning "value of $\delta$ at $x$ equals value of $\delta$ at $-x$" is meaningless/invalid.

But statement $$ \int dx~ \delta(x) f(x) = \int dx~\delta(-x) f(x) \quad \text{for all functions }f \tag{**} $$ may be valid.

You can easily verify that the function of $\Delta$ and $x$ ( the expression after the limit sign in definition of $\xi$) does not satisfy either of these two statements (in the role of $\delta$). So it is not "symmetric".

The delta distribution can hypothetically satisfy only the second statement. Does it do so?

We can evaluate both sides of the equality. The left-hand side has value, by definition of $\delta(x)$, $f(0)$.

We can transform the right-hand side integral into $$ \int dx~\delta(-x) f(x) = \int dy~\delta(y) f(-y) $$ By definition of $\delta(y)$,value of this integral is $f(0)$, the same as the left-hand side. So (**) is satisfied.

The equation $\delta(x) = \delta(-x)$ is thus consequence of the definition of $\delta(x)$, it is not independent assumption.

Your function $\xi$ may actually obey the second statement too (and thus be symmetric in that sense), even though the $\Delta$-dependent expression after the limit sign does not. This is similar for other approximations of delta distribution; the approximation may not have properties of $\delta$ (such as symmetry), but the limit does.

The symbol $$\delta(x\!-\!y)\tag{A}$$ with two arguments $x,y\in\mathbb{R}$ is an informal kernel notation for the Dirac delta distribution $$u~\in~ D^{\prime}(\mathbb{R}^2)\tag{B}$$ defined as

$$u[f]~:=\int_{\mathbb{R}}\!\mathrm{d}z~f(z,z)\tag{C}$$

for testfunctions $$f~\in~ D(\mathbb{R}^2).\tag{D}$$ It follows that the Dirac delta defined as above is symmetric $$ \delta(x\!-\!y)~=~\delta(y\!-\!x), \tag{E}$$ cf. OP's title question.