$\def\abs#1{\left|#1\right|}$I came across this lemma in my Real Analysis textbook, but I'm not sure exactly what it is saying. Here it is:
Lemma:Let $g: D \rightarrow \mathbb{R}$ be continuous at $x_{0} \in D$ with $g(x_{0}) \neq 0$. Then there are $\delta >0$ and $\alpha >0$ such that if $\abs{ x-x_{0}} < \delta$ and $x \in D$, then $\abs{g(x)} \geq \alpha $.
I can supply the proof if requested. The lemma comes about in trying to prove that if $f$ and $g$ are both continuous and $g(x) \neq 0$ , then $f/g$ is continuous at $x_{0}$.