In Wikipedia page on germ, it says that
Given a point $x$ of a topological space $X$, and two maps $f,g\,:\,X\to Y$ (where $Y$ is any set), then $f$ and $g$ define the same germ at $x$ if there is a neighbourhood $U$ of $x$ such that restricted to $U$, $f$ and $g$ are equal.
My question is: given two functions $f,g$ in $\mathbb{R}\times \mathbb{R}$, if $D_x f|_{x=0} = D_x g|_{x=0}$ and $f(0) = g(0)$, do they define the same germ at $x=0$?
Attempts to solve this question: If something like $U = \{\epsilon\,|\,|\epsilon|\to 0^+\}$ is an eligible neighbourhood, $D_x f|_{x=0} = D_x g|_{x=0}$ and $f(0) = g(0)$, then $\forall x\in U$, $f(x) = g(x)$, and the answer is affirmitive. However, this is not rigorous and I didn't find examples of this kind of ''limiting-case'' neighborhoods. Any guidance on this question is greatly appreciated!
Edit: thanks to the comment and answer, this is not true. As I understand it, $U = \{x\,|\,x\in \mathbb{R}, \forall \epsilon>0, |x|<\epsilon\}$ is the set $\{0\}$ and does not include an openset, so not a neighborhood.