Continuous function positive at a point is positive in a neighborhood of that point

Pretty much the problem asks if a function is continuous at the point $c$ and $f(c) > 0$ then there exists a $d > 0$ such that $\forall x$, $f(x) > 0$ with $|x-c| < d$.

I can understand what the problem means. That if a function is positive at a point then there is another point that's really close that'll also be positive. I can not prove this in a formal way though. I've tried using the intermediate value theorem and I do not know hot to implement it. Any ideas?

Solution 1:

This will follow from the definition of continuity applied to the point $x=c$. In particular, $\exists \delta > 0$ such that for any $\epsilon > 0$:

$|f(c) - f(x)| < \epsilon \, \, \, \, \mbox{if} \, \, \, \, |x-c| < \delta$.

This is equivalent to saying that $f(x)$ is within $\epsilon$ of $f(c)$:

$f(x) \in (f(c) - \epsilon, f(c) + \epsilon)\, \, \, \, \mbox{if} \, \, \, \, |x-c| < \delta$.

So that clearly $f(x) > f(c) -\epsilon$. Thus, simply choose any value for $\epsilon$ for which $0<\epsilon < f(c)$ (which we can do because $f(c)$ is positive by assumption). Thus, $\exists \delta$

$f(x) > f(c) - \epsilon > 0 \, \, \, \, \mbox{if} \, \, \, \, |x-c| < \delta$.

Solution 2:

Hint: Consider the definition of continuity.

$f$ is continuous at $x_0$ if and only if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|x - x_0| < \delta$ implies that $|f(x) - f(x_0)| < \epsilon$.

If $f$ is continuous, then what does continuity imply when $|f(x) - f(x_0)| < f(x_0)$?