f a function from (a,b) to reals and is continuous at c in (a,b) where f(c) is not 0. Show there exists a d>0 st |f(x)|>|f(c)|/2 for all x st |x-c|$f$ a function from $(a,b)$ to reals and is continuous at $c$ in $(a,b)$ where $f(c)$ is not $0$. Show there exists a $d>0$ st $|f(x)|>|f(c)|/2$ for all $x$ st $|x-c|<d$.
I've been playing around with the fact that $|x-y|\geq ||x|-|y||$ and the preservation of sign rule, but can't get anywhere.
Any help would be appreciated.

Let $\varepsilon = \frac{|f(c)|}{2}$. Since $f(c) \neq 0$, $\varepsilon > 0$. Thus by continuity of $f$ at $c$, there exists a $d > 0$ such that $|f(x) - f(c)| < \varepsilon$ for all $x$ such that $|x - c| < d$. Use the triangle inequality to show that $|f(x)| > |f(c)| - \varepsilon$ whenever $|x - c| < d$.
