In one of my homework assignments (in intro to applied mathematics) there is the following definition:
Given two functions $f(\epsilon),g(\epsilon)$ that are defined in $D=(0,\epsilon)$ we say that $f(\epsilon)=o(g(\epsilon))$ if there exist a bounded function $c(\epsilon)$ defined for $\epsilon\in D$ s.t $$|f(\epsilon)|<c(\epsilon)|g(\epsilon)|$$ for $\epsilon\in D$.
This definition is a bit confusing to me:
First, why is there an $\epsilon$ in the definition of $D$ ? isn't $\epsilon$ just a dummy variable and can be any number ?
Second, taking $c=2$ and $$g(\epsilon)=f(\epsilon)=\epsilon$$ we get that $f=o(f)$. But I remember that this notation can also be defined by means that the limit $$\lim_{\epsilon\to0}\frac{f}{g}=0$$ (or maybe I am just confusing this with the notation of $o$ that I know for functions $f(n)$ where $n$ is a positive integer).
Third, It seems that this definition is the same as saying that there exist a constant $c$ s.t $$|f(\epsilon)|<c|g(\epsilon)|$$ since we took $c(\epsilon)$ to be bounded denote $$c=\sup_{\epsilon\in D}c(\epsilon)$$ But the assignments defines $f(\epsilon)=O(g(\epsilon))$ if there exist a constant $c$ s.t $|f(\epsilon)|<c|g(\epsilon)|$ for $\epsilon\in D$ so either I'm understanding wrong, or $o$ and $O$ means the same.
Is there a problem in the definitions given me (i.e whomever wrote them probably meant something else) or am I understanding wrong the definition ?