We consider the closed subscheme $X:= V(Y^2-X^2(x+1))$ of affine plane $\mathbb{A}^2_k= Spec \text{ } k[X,Y]$. field $k$ is arbitrary. let $x:=(0,0)$ the point representing the prime ideal $(X,Y) \subset k[X,Y]$.
Q: how to prove that the completion $\widehat{\mathcal{O}}_{X,x}$ of the stalk $\mathcal{O}_{X,x}$ at $x \in X$ is not a domain although $X$ itself and $\mathcal{O}_{X,x}$ are integral.
the motivation for this question comes from a comment in Wedhorn's & Goetz' Algebraic geometry book and motivates the phenomena that the coarseness of Zariski topology often "hides" infinitesimal anomal behavior that violate certain theorems impose local-global principles which hold when we work with Zariski topology.
Back to my question. I'm still not familar with work with inverse limits & it's features. by definition the completion of $\mathcal{O}_{X,x}$ wrt $x=(0,0)$ is
$$\widehat{\mathcal{O}}_{X,x}= \varprojlim_{n\in \mathbb {N} } \mathcal{O}_{X,x}/\mathfrak{m}^n $$
with $\mathfrak{m}:=(y,x)$. by assumption $\mathcal{O}_{X,x}$ is the localization of $k[X,Y]/(Y^2-X^2(x+1))$ at $\mathfrak{m}$.
I see two ways to reach my goal: one is to construct a compatible non zero chain $r:=(a_0,a_1, ..., a_i, ...)$ with $a_i \in \mathcal{O}_{X,x}/\mathfrak{m}^i$ such that $a_{i+1} +\mathfrak{m}^i=a_i$ and there exist another non zero compatible chain $s:=(b_0,a_1, ..., b_i, ...)$ with $a_j \cdot b_j \in \mathfrak{m}^j$ for every $j \in \mathbb{N}$. this mean that $r,s$ represent two elements in $\widehat{\mathcal{O}}_{X,x}$ and are zero divisors.
another idea is just to calculate $\widehat{\mathcal{O}}_{X,x}$ directly. is it in this case a serious option or too difficult? is there a standard way to calculate completions of local rings with structure $k[X,Y]_{\mathfrak{p}}/(f)$ for $f \in k[X,Y]$ and $\mathfrak{p}$ prime containing f.