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I know that there are a lot of similar questions like this on this forum but still I can't figure it out one thing of this definition.

Definition: f is continuous at $x_o\in X\subset\mathbb{R}$ if

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

My question is, why in this definition I can't take $\epsilon \to 0$?

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Please try to remember that there is no real number at all that tends to zero. You can't write, in standard analysis, anything like "consider a number $x \to 0$."

The very definition of limit actually gives the piece of notation "$f(x) \to L$ as $x \to x_0$ " a meaning by using quantifiers: for every $\epsilon>0$ there exists $\delta>0$ etc. So, to summarize: you can arbitrarily pick a positive number, but you can't let real numbers move towards a limit value.

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  • $\begingroup$ Thank you, little doubt. Can I take, for example, $\epsilon=\epsilon_n=\frac{1}{n}$ (which is, $\forall n$ , a real number) ? If i consider it for $n\to\infty$, it goes to zero. Why this doesn't hold? $\endgroup$ – Django F. Dec 22 '17 at 16:24
  • $\begingroup$ Well, it does work! $\endgroup$ – Siminore Dec 22 '17 at 16:25
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The idea is that if the inequality on the right holds for any $\epsilon$, then it holds for "really small ones". That is we can make $\epsilon$ as close to zero as we want and if the inequality holds no matter what tolerance you use, the two must be equal.

So, it seems like you have the right idea by thinking about $\epsilon$ being really small.

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