Question:
Suppose that $f$ is a continuous function at the point $x_0$ and that $f(x_0) \neq 0$. Using the $\epsilon−\delta$ definition of the continuity of a function at $x_0$, show that there exists $\delta>0$ such that $$f(x_0) \neq 0 \ {whenever}\ |x-x_0| < \delta \ and \ x\in[a,b] $$
My thoughts are we use the defintion as stated in the question :
$|f(x) - f(x_0)| < \epsilon$ whenever $|x-x_0| < \delta$ and then we can use the triangle inequality "$|a-b|| \geq ||a| - |b||$" and then use the fact that $f(x_0) \neq 0 $ and then when we balance the inequality we get that $\epsilon - f(x_0) \neq 0 $ which we can say that it is $\neq 0$
If someone could guide me through how to complete this quesition to the end, it would be really appreciated, thank you!
Suppose WLOG that $f(x_0)\lt 0$ so that $-f(x_0)\gt 0$
Since $f$ is continuous at $x=x_0$, it follows that for every $\epsilon\gt 0,\exists \delta_\epsilon\gt 0$ such that $|x-x_0|\lt \delta_\epsilon\implies |f(x)-f(x_0)|\lt \epsilon$. Choose $\epsilon = -f(x_0)$ and hence by continuity of $f$ at $x_0$, there exists a $\delta\gt 0$ such that
$\begin{align}|x-x_0|\lt \delta &\implies |f(x)-f(x_0)|\lt -f(x_0)\\ &\implies f(x_0)\lt f(x)-f(x_0)\lt -f(x_0)\\ &\implies 2f(x_0)\lt f(x)\lt 0\\ &\implies \text{$f(x)\ne0$ for $x:|x-x_0|\lt \delta$}\end{align}$
Similarly, prove for $f(x_0)\gt 0$.