I am new to the mechanics of the $\epsilon$-$\delta$ definition of continuity of a function at a point, and so I am having trouble selecting my $\delta$ for the following question:
Suppose a function $f$ is continuous at a point $c$ and $f(c) > 0$. Prove that there is a $\delta > 0$ such that for all $x \in \mathrm{domain}(f)$, $ |x - c| \le \delta \ \Rightarrow \ f(x) \ge \frac{f(c)}{2} $
I tried starting with the definition of continuity i.e.
If $f$ is continuous at $c$ then letting $\epsilon > 0$ be given there exists a $\delta > 0$ such that $|x -c| \le \delta$ implies $|f(x) - f(c)| \le \epsilon$
However, I haven't found a way to work in the condition that $f(c) > 0$. Should I try a contrapositive proof?
Edit: I see! Thank you. So this is saying that there is always a $\delta$ such that $f(x)$ is no more than $f(c)/2$ away from $f(c)$ and as the answerer pointed out, indeed any distance $z$ away from the limit.