Continuous of function. If $f$ continuous at $x_0$ and $f(x_0)>0$ 
then how to show there exist $\epsilon>0$ and $\delta>0$ such that $f(x)>0$ for $|x-x_0|<\delta$.
I don't know to start proving and then i use definition continuous function at $x_0$ is
$\forall\epsilon>0, \, \exists\delta>0 \, \text{such that if }|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$ 
 A: Since $f(x_0)>0$ so is $\frac{f(x_0)}{2}>0$. Let $\epsilon = \frac{f(x_0)}{2}$. By continuity of $f$ we know that there exists $\delta>0$ such that,
$$|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon = \frac{f(x_0)}{2}\\
\implies -\frac{f(x_0)}{2}<f(x)-f(x_0)<\frac{f(x_0)}{2},\\
\implies \frac{f(x_0)}{2}<f(x)<\frac{3f(x_0)}{2}$$
since both sides are positive, so is $f(x)$, the end.
A: Pick $\varepsilon = f(x_0) > 0$. Then, by the definition of continuity that you wrote, we have:
For $\varepsilon = f(x_0) > 0, \, \exists\delta>0$ such that if $|x-x_0|<\delta$, then:
$$|f(x) - f(x_0)|<\varepsilon = f(x_0) \quad \quad (\iff -\varepsilon < f(x) - f(x_0) < \varepsilon)$$
so
$$-f(x_0) < f(x) - f(x_0) < f(x_0) \implies 0 <f(x) < 2f(x_0) $$
A: $f$ continuos at $x_0$, and $f(x_0) >0$.
For every $\epsilon >0$ there is a $\delta >0$ such that
$|x-x_0| \lt \delta$ implies 
$|f(x)-f(x_0)| \lt \epsilon.$
Since $|f(x)-f(x_0)| \lt \epsilon$, we have
$-\epsilon \lt f(x)-f(x_0) \lt \epsilon$, or
$f(x_0) -\epsilon \lt f(x)$.
For which choice of $\epsilon$ is the LHS positive?
Recall: $f(x_0) >0.$
