Suppose $f$ is a "nice" function. Show that if $f$ is continuous and with $f(p)\neq 0$ , then for some $\delta$ neighbourhood of $p$ we have: $$|f(x)| > \frac{1}{2} |f(p)|$$
I can obtain $f(x) > \frac{1}{2} f(p)$ quite easily from the triangle inequality but unsure of how to get the inequality with modulus included.
EDIT: Just realised something, we have: $$|f(x) - f(p) | \rightarrow 0$$ But in particular this implies $$|~~ |f(x)| - |f(p)| ~~| \rightarrow 0$$ And then the desired inequality drops out by setting $\epsilon = \frac{ |f(p)|}{2}$. Thank you for the help in the commenets :)