Intuitive explanation for $\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$

Here is my approach. Maybe not the best way since it's Weierstrass product of Gamma function. Anyway, we have: $$\Gamma (z+1) = e^{-\gamma z} \prod_{k=1}^{\infty} \left(1+\frac{z}{n}\right)^{-1} e^{z/n}\Rightarrow \log \Gamma (z+1)=- \gamma z + \sum_{k=1}^{\infty}\left[\frac{z}{k}- \ln \left(1+\frac{z}{k}\right)\right] $$ Now let $a_n:= n\left[2\log \Gamma (n+1) - \log \Gamma (n+1-x) - \log \Gamma (n+1+x)\right]$. Which we need to prove the fact that $\lim_{n \to \infty} e^{a_n}= e^{-x^2}$ or prove $ \lim_{n \to \infty} a_n=-x^2$. Doing some algebraic we obtain: $$a_n = n \cdot \sum_{k=1}^{\infty}\left[\ln\left(1 + \frac{n+x}{k}\right)+\ln\left(1 + \frac{n-x}{k}\right) - 2\ln\left(1 + \frac{n}{k}\right)\right]$$$$=n \cdot \sum_{k=1}^{\infty} \ln\left(1- \frac{x^2}{(n+k)^2}\right)=-~x^2 \cdot \sum_{k=1}^{\infty} \frac{n}{(n+k)^2}+\mathcal{O}(x^4)\cdot \sum_{k=1}^{\infty} \frac{n}{(n+k)^4} $$ It is obvious by comparison test that $\cdot \sum_{k=1}^{\infty} \frac{n}{(n+k)^4}\rightarrow 0, n \to \infty $. For the first sum, by Riemann's sum we obtain: $$\lim_{n \to \infty}\sum_{k=1}^{\infty} \frac{n}{(n+k)^2} = \lim_{n \to \infty}\frac{1}{n} \cdot \sum_{k=1}^{\infty} \frac{1}{(\frac{k}{n}+1)^2} = \int_{0}^{\infty} \frac{1}{(x+1)^2}\mathrm{d}x = 1$$ This result yields that $\lim_{n \to \infty} a_n = - x^2$. $\blacksquare$

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.