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Problem:

Let $\, f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous at the point $x_0$ and $f(x_0) \ne 0 \ \forall x \in \mathbb{R}$. Show that $\exists \delta > 0$ such that $\vert f(x) \vert \geq \ \frac{\vert f(x_0) \vert}{2}>0 \ \forall x \in \mathbb{R}$ for $\vert x-x_0 \vert < \delta $

Attempted solution:

My idea so far has been to find $\epsilon > 0$ so that I could derive the inequality from $\vert f(x)-f(x_0) \vert < \epsilon \ \forall x \in \mathbb{R}, \vert x-x_0 \vert < \delta$, where the existence of such $\delta$ is guaranteed by the definition of continuity.

I tried quite a few different values of $\epsilon$, one of which was $\epsilon=\vert f(x_0) \vert >0$. Using this value, I arrived at $\frac{\vert f(x)-f(x_0) \vert}{\vert f(x_0) \vert}<1 \iff 0<f(x)<2f(x_0) $ but this isn't helpful at all. Any ideas?

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    $\begingroup$ Then how about $\varepsilon = |f(x_0)|/2$? $\endgroup$ – xbh Nov 12 '18 at 18:00
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As xbh said, we can find $\delta>0$ such that $$|f(x)-f(x_0)|<\frac{|f(x_0)|}{2} \hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta) $$ since $f$ is continuous at $x_0$ and $f(x_0)\neq 0$ so $|f(x_0)|>0$. Then by triangular inequaltiy, $$|f(x_0)|-|f(x)|\leq|f(x)-f(x_0)|<\frac{|f(x_0)|}{2}\hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta) $$ so, $$\frac{|f(x_0)|}{2}=|f(x_0)|-\frac{|f(x_0)|}{2}< |f(x)| \hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta). $$

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