$\epsilon$-$\delta$ problem (continuous function)

Question

Question:
Let f : I → R be a continuous function on the interval I, and let $f(a) > 0$ for some $a\in I.$

Then,Show that there exists some $δ > 0$ and that $f(x) > 0$ for some $x\in I$ satisfying $|x - a| < δ$.

I tried to prove using
$\displaystyle\lim_{x\to a}{f(x)}=f(a) \iff(\forall\epsilon>0)(\exists\delta>0)(\forall a\in I):|x-a|<\delta\Rightarrow|f(x)-f(a)|<\epsilon$
and
$|f(x)-0|\leq|f(x)-f(a)|+|f(a)-0|$
but,I can’t find a good solution.

Thanks,for any help

Answer 1

Since $\frac{f(a)}2\gt 0$, by continuity of $f$ at $a$, there exists a $\delta \gt 0$ such that $|x-a|\lt \delta \implies |f(x)-f(a)|\lt \frac{f(a)}2\implies \frac{f(a)}2\lt f(x)\lt \frac{3f(a)}2\implies f(x)\gt 0 $ for all $x$ such that $|x-a|\lt \delta$.