Central limit theorem for discrete random variable

 Suppose that $X'$ is a random variable with $\Pr(X'=1)=\Pr(X'=-1)=0.5$. Then $$X=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}(X_1+X_2+\dots+X_n)$$ approaches normal random variable with zero mean and unity variance, where $X_i$s are independent samples of $X'$.

Below we prove that $$\Pr(X=x)\approx e^{-\frac{x^2}{2}}\Pr(X=0).$$

==================

Proof:

We consider $2n$ samples instead. $X=x$ translates to $n+k$ samples of $1$ and $n-k$ samples of $-1$ among $X_1,\dots,X_n$ where $$k=\frac{x\sqrt{2n}}{2}.$$ For any $k\in\mathbb{N}$ we have $$\frac{\Pr(X=x)}{\Pr(X=0)}=\frac{\binom{2n}{n+k}}{\binom{2n}{2n}}=\frac{n!n!}{(n+k)!(n-k)!}=\frac{n(n-1)\dots(n-k+1)}{(n+k)(n+k-1)\dots(n+1)}\approx(1-\frac{k}{n})^k.$$ Substituting $x\sqrt{2n}/2$ for $k$ we get $$\frac{\Pr(X=x)}{\Pr(X=0)}\approx\left(1-\frac{x}{\sqrt{2n}}\right)^{\frac{\sqrt{2nx}}{2}}\approx e^{-\frac{x^2}{2}}.$$