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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a real function. $f$ is continuous at point $c$ iff

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(|x-c|<\delta\Rightarrow|f(x)-f(c)|<\epsilon)$$

Continuity is also defined at a point $c$ if $\lim\limits_{x\rightarrow c}f(x)=f(c)$ i.e.

$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(0<|x-c|<\delta\Rightarrow|f(x)-f(c)|<\epsilon)$$

So this is a trivial question but how do I show these definitions are equivalent? They are almost the same except that in the limit definition we have $0<|x-c|<\delta$ So how can I get rid of it in this case. As I said it is trivial and one can see it but can't see how to formally show it, so any help please.

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  • $\begingroup$ In the second definition, you can relax the $|x-c|>0$ condition because when $x=c$, $f(x)=f(c)=L$ is equal to the limit. $\endgroup$ – user142299 May 19 '14 at 15:11
  • $\begingroup$ Yes thanks I see that but my question was how to show it, so starting with 1st definition I want an iff argument, that is definition 1 iff ... iff... iff definition 2. Is there such an argument or is it just the case we relax the condition as x=c is allowed here so its just that obvious and I am complicating things? Feedback greatly appreciated $\endgroup$ – Sam May 19 '14 at 15:24
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Take the first definition as given. Then clearly the second condition follows, since the range of $x$ values being considered is a subset of the range provided by the first definition.

If we take the second definition as given, we have to show that the weaker condition $$0<|x-c|<\delta\implies |f(x)-f(c)|<\epsilon$$ implies the stronger one $$|x-c|<\delta\implies |f(x)-f(c)|<\epsilon$$ So we need only verify that for $x=c$, we still have $|f(x)-f(c)|<\epsilon$. But this is easy, because when $x=c$, $|f(x)-f(c)|=|f(c)-f(c)|=0<\epsilon$ regardless of the value of $\epsilon>0$.

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