Let $f : \mathbb{R} \to \mathbb{R}$ be continuous at $c$ and let $f(c) > 0$. Show that there exists an open interval $I$ containing $c$ such that $f(x) > 0$ for all $x \in I$.
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In the $\delta-\epsilon$ formulation, you establish that $|f(x)-f(c)|<\delta$, don't you ? – Dec 01 '16 at 15:06
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Hint. Recall that continuity implies that $\lim_{x\to c}f(x)=f(c)$. Then by the definition of limit, $$\forall \epsilon>0,\exists\delta>0: |x-c|<\delta \implies |f(x)-f(c)|<\epsilon.$$ This means that, for all $x\in I:=(c-\delta,c+\delta)$, $$f(c)-\epsilon<f(x)<f(c)+\epsilon.$$ Choose $0<\epsilon<f(c)$. What may we conclude?
Robert Z
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Yes thus it follows that f(x) is greater than 0, but this is only for one value of epsilon is that allowed? – Ayush Kumar Dec 01 '16 at 14:45
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1@Ayush Kumar Fixing $\epsilon$ is fine. You are looking for $x$ such that $f(x)>0$. This holds for all $x$ such that $|x-c|<\delta$ that is for $x\in (c-\delta,c+\delta)$. – Robert Z Dec 01 '16 at 14:54
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