To prove:
Given that $\lim_{x\to c} g(x) = L$, where $L > 0$, prove that there exists an open interval $(a,b)$ containing $c$ such that $g(x) > 0, \forall x \neq c$ in $(a,b)$.
My work so far:
Since we are given that $\lim_{x\to c} g(x) = L$ we know that $\forall \varepsilon > 0, \exists \delta$ such that $0<\vert x - c \vert < \delta \Rightarrow \vert g(x) - L \vert < \varepsilon$.
Now, it seems that we should be able to use the information directly above this sentence, as well as the fact that $L > 0$ to show that such an interval exists. Obviously our values $g(x)$ are approaching some positive value and thus (even if our interval is incredibly small), there must be positive values arbitrarily close to $L$ on both sides of $L$.
I can't quite figure out how to formalize this thought in a proof. Can anyone offer some hints and/or let me know if I'm on the right track in my line of thinking?