Intuitive explanation for $\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$ In this post I noticed (at first numerically) that:
$$\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$$
This can be proved by looking at the Taylor expansion
$$n\ln\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)=-2n\sum_{k=1}^{\infty}\frac{\psi^{(2k-1)}\left(n+1\right)}{\left(2k\right)!}x^{2k}$$
and the asymptotic expansion
$$\psi^{(m)}(n+1)=\left(-1\right)^{\left(m+1\right)}\sum_{k=0}^{\infty}\frac{\left(k+m-1\right)!}{k!}\frac{B_k}{n^{k+m}}$$
where we have chosen $B_1=-\frac12$.
However, this limit seems so beautiful and interesting that it produces the Gaussian function. It makes me wonder if there is a more intuitive way to understand this limit, possibly in a context of probabilities.

Consider first
$$ \begin{align}
\frac{(n+5)!}{n!} &=(n+5)(n+4)\cdots (n+1)\\
&= n^5 (1+5/n)(1+4/n) \cdots (1+1/n) \\
& \approx n^5 \left(1 + \frac{5+4+\cdots+1}{n}\right) \tag1\\
& = n^5 \left(1 + \frac{5\times6}{2n}\right) \\
\end{align}$$
Or, in general
$$ \frac{(n+x)!}{n!} \approx n^x \left(1 + \frac{x(x+1)}{2n}\right) \tag2$$
Similarly:
$$ \frac{n!}{(n-x)!} \approx n^x \left(1 - \frac{x(x-1)}{2n}\right) \tag3$$
Then
$$
\begin{align}
\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}
&\approx \frac{\left(1 - \frac{x(x-1)}{2n}\right)}{\left(1 + \frac{x(x+1)}{2n}\right)}\\
&\approx \left(1 - \frac{x(x-1)}{2n}\right) \left(1 - \frac{x(x+1)}{2n}\right)\\
&\approx \left(1 - \frac{x^2}{n}\right) \tag4
\end{align}
$$
And, using the standard exponential limit:
$$\lim_{n\to \infty} \left(1 - \frac{x^2}{n}\right)^n =e^{-x^2} \tag5$$
Edit Result $(4)$ can also be obtained by a probabilistic reasoning: Imagine the following scenario: from a bag with $n$ white balls and $x$ black balls we pick $x$ balls. Which is the probability that all picked balls are white?
This is $$ \frac{\binom{n}{x}}{\binom{n+x}{x}}=\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}$$
precisely the LHS in eq $(4)$.
Now, if $n \gg x$, the probability that more than one of the picked balls are black is negligible, hence we can approximate this by $P \approx 1 - x \frac{x}{n}= 1- x^2/n$
BTW: The complete term in the original limit, $\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}$, can then be regarded as the probability of getting no black balls (after adding $n$ white balls) in $n$ tries.
I don't see how this probability could be associated with a gaussian density.
