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We know that the definition of a limit is $\forall\epsilon > 0 \space \exists\delta >0, 0 < |x-c| < \delta \space \to \space |f(x) - L| < \epsilon$

However, won't the definition stay the same if we were to write $\forall\epsilon > 0 \space \exists\delta \ge0, \space0 < |x-c| < \delta \space \to \space |f(x) - L| < \epsilon$, since for each epsilon there has to be some delta that works. Therfore, if since $\delta \ge 0$, it follows that $\delta > 0$?

Also, shouldn't that be the same for $\forall\epsilon > 0 \space \exists\delta >0, 0 < |x-c| < \delta \space \to \space |f(x) - L| \le \epsilon$ as well?

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    If $\delta = 0$, then $0 < |x - c| < \delta$ is always false, so the overall implication is always true. So the definition would lose all content. –  Oct 04 '17 at 04:23
  • But we don't have to pick $\delta = 0$ do we? Don't we just pick a delta that works? –  Oct 04 '17 at 04:24
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    The point is that if you can choose $\delta = 0$, the implication is always true. Since it's bound by an existential quantifier $\exists$, this is a real problem. –  Oct 04 '17 at 04:25
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    The first $<$ matters, the $<$ for the $|f(x)-L|$ doesn't. – copper.hat Oct 04 '17 at 04:31
  • @copper.hat then why don't we just write $|f(x) - L| < \epsilon$ –  Oct 04 '17 at 04:35
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    Do you mean as opposed to $|f(x)-L| \le \epsilon$? – copper.hat Oct 04 '17 at 04:35
  • sorry, I mean to say why dont we write it as $|f(x) - L| \le \epsilon$ as opposed to $|f(x) - L| < \epsilon$ –  Oct 04 '17 at 04:39

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You are allowed to write $\exists\delta \ge0$ instead of $\exists\delta>0$ since you see in the definition of the limit that $\delta$ cannot take the value $0$ because that would produce contradiction in definition.

You shuldn´t write $\epsilon \geq 0$ instead of $\epsilon >0$ since for $\epsilon =0$ you would have that $f$ is necessarily constant function and you want that our definition of the limit covers more general class of functions than just constant functions.