Continuous of function.

Question

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

$\forall\epsilon>0, \, \exists\delta>0 \, \text{such that if }|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$

Answer 1

$f$ 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: $f(x_0) >0.$

Answer 2

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.

Answer 3

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

$$-f(x_0) < f(x) - f(x_0) < f(x_0) \implies 0 <f(x) < 2f(x_0) $$