Continuous of function. If $f$ continuous at $x_0$ and $f(x_0)&gt;0$ 

then how to show there exist $\epsilon&gt;0$ and $\delta&gt;0$ such that $f(x)&gt;0$ for $|x-x_0|&lt;\delta$.

I don't know to start proving and then i use definition continuous function at $x_0$ is

$\forall\epsilon&gt;0, \, \exists\delta&gt;0 \, \text{such that if }|x-x_0|&lt;\delta\implies |f(x)-f(x_0)|&lt;\epsilon$
Since $f(x_0)&gt;0$ so is $\frac{f(x_0)}{2}&gt;0$. Let $\epsilon = \frac{f(x_0)}{2}$. By continuity of $f$ we know that there exists $\delta&gt;0$ such that,

$$|x-x_0|&lt;\delta\implies |f(x)-f(x_0)|&lt;\epsilon = \frac{f(x_0)}{2}\\
\implies -\frac{f(x_0)}{2}&lt;f(x)-f(x_0)&lt;\frac{f(x_0)}{2},\\
\implies \frac{f(x_0)}{2}&lt;f(x)&lt;\frac{3f(x_0)}{2}$$

since both sides are positive, so is $f(x)$, the end.
$f$ continuos at $x_0$, and $f(x_0) &gt;0$.

For every $\epsilon &gt;0$ there is a $\delta &gt;0$ 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: $f(x_0) &gt;0.$