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I am new to the mechanics of the $\epsilon$-$\delta$ definition of continuity of a function at a point, and so I am having trouble selecting my $\delta$ for the following question:

Suppose a function $f$ is continuous at a point $c$ and $f(c) > 0$. Prove that there is a $\delta > 0$ such that for all $x \in \mathrm{domain}(f)$, $$ |x - c| \le \delta \ \Rightarrow \ f(x) \ge \frac{f(c)}{2} $$

I tried starting with the definition of continuity i.e.

If $f$ is continuous at $c$ then letting $\epsilon > 0$ be given there exists a $\delta > 0$ such that $|x -c| \le \delta$ implies $|f(x) - f(c)| \le \epsilon$

However, I haven't found a way to work in the condition that $f(c) > 0$. Should I try a contrapositive proof?


Edit: I see! Thank you. So this is saying that there is always a $\delta$ such that $f(x)$ is no more than $f(c)/2$ away from $f(c)$ and as the answerer pointed out, indeed any distance $z$ away from the limit.

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1  
Have you sketched a continuous curve in the Cartesian plane for inspiration? That can help you visualize if you are bogged down in symbols. –  rschwieb Aug 16 '12 at 19:26
    
I should point out that you are given that $f$ is continuous at $c$. You do not choose a value for $\delta$ (as in the definition), but rather a value of $\epsilon$. The definition then furnishes the $\delta$ for you. You need to choose $\epsilon$ small enough so that your condition is satisfied. –  David Mitra Aug 16 '12 at 19:36
    
I see your point. I misspoke, evidently out of a bit of confusion with the definition. –  Mike Aug 16 '12 at 19:38
    
I changed "$\epsilon-\delta$" to "$\epsilon$-$\delta$". When you put the hyphen INSIDE the TeX setting, then it looks like a minus sign instead of a hyphen. –  Michael Hardy Aug 16 '12 at 20:23

2 Answers 2

up vote 2 down vote accepted

Let $\epsilon = \frac{f(c)}{2}$. Since $f(c) > 0$, you have that $\epsilon > 0$. By continuity, there exists a $\delta$ such that for all $x$ such that $|x - c| < \delta$, you have that $|f(x) - f(c)| < \epsilon$. This implies that

$f(c) - \epsilon < f(x) < f(c) + \epsilon$

we only care about the left side. Remembering that $\epsilon = \frac{f(c)}{2}$, you get

$$f(c) - \frac{f(c)}{2} = \frac{f(c)}{2} < f(x).$$

for $|x - c| < \delta$, as desired.

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Let me suggest a two-steps proof:

  • Use the fact that $f$ is continuous at $c$ to show that, for every $z\lt f(c)$, there exists a $\delta > 0$ such that $|x -c| \leqslant \delta$ implies $f(x)\geqslant z$.
  • Use the fact that $f(c)\gt0$ to show that $z=\frac12f(c)$ is such that $z\lt f(c)$.
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