Let $\mathbb{F}$ be an algebraically closed field and $X$ an affine algebraic set. Fix $x \in X$. For an open set $U \subset X$ let $\mathcal{O}_X(U)$ denote the collection of regular functions on $U$. Given two open neighborhoods $U \subset V$ of $x$ then restriction defines a map of algebras $\mathcal{O}_X(V) \to \mathcal{O}_X(U)$. The family of algebras $\mathcal{O}_X(U)$ as $U$ runs over all open neighborhoods of $x$ is an inverse system with respect to these restriction maps. I want to show that $\mathcal{O}_x:= \mathbb{F}[X]_{M_x}$ is the inverse limit of this inverse system in the category of $\mathbb{F}$ algebras. Here $M_x \subset F[X]$ is the maximal ideal of functions which vanish at $x$.
In order to show $\mathcal{O}_x$ is this inverse limit I want to find $\pi$, a collection of maps $\{\pi_U : U \subset X \text{ an open neighborhood of }x\}$ such that for each open set $U$ we have $\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$. Further I need $(\mathcal{O}_x,\pi)$ to be a cone over my inverse system. However I'm struggling even to define a map $\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$. It seems that it really suffices to define $\pi_U$ for $U = X$ where $\mathcal{O}_X(X) = \mathbb{F}[X]$ as, for $(\mathcal{O}_x,\pi)$ to be a cone we must have $\pi_U\left(\frac{f}{g}\right) = \pi_X\left(\frac{f}{g}\right)\bigg{|}_{U}$ for any open set $U$.
After thinking about this for a bit, I think I might be misinterpreting something but I'm really not sure where. Any help would be appreciated.