Let $D\subseteq\Bbb R$ and $a\in D$. If $f:D\to\Bbb R$ is "$\epsilon$-$\delta$" continuous at $a$ and $f(a)\neq0$, prove that there exists a $\delta\in\Bbb R_{>0}$ such that for all $x\in D$ we have $|x-a|<\delta\Rightarrow\dfrac{|f(a)|}2<|f(x)|$.
From the definition of continuity, I know that for every $\epsilon$ $\gt{0}$, I must get a $\delta$ $\gt{0}$ In this case, I note that:
$\vert x-a \vert$ $\lt$ $\delta$ $\Rightarrow$ $\frac{|f(a)|}{2} - \epsilon$ $\lt$ $|f(x)| \lt \frac{|f(a)|}{2} + \epsilon$
Which tells me that
$\frac{|f(a)|}{2} - |f(x)| \lt \epsilon$
I am confused as how could I find a $\delta$ from here. Does anyone have a hint? Am I even going the right direction? Any help is appreciated.
Simply take $\epsilon= \frac{|f(a)|}{2}$ which by hypothesis is a positive number.