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To prove that $\lim_{x\to a}f(x) = L$, we need to show that for any $\epsilon>0$, there exists a $\delta>0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

What's the reasons for epsilon and delta have sing $<$ not a $\leq$ ? And Why do we assume for epsilon and some delta? Not either for all epsilon and all delta or some epsilon and all delta etc..

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  • $\begingroup$ The definition is also equivalent if you take $\leq \varepsilon$. Assume the original definition is satisfied. Then the alternate definition is also satisfied, since $|f(x) - L| < \varepsilon \implies |f(x) - L| \le \varepsilon$. And the alternate definition implies the original definition: pick any $\varepsilon > 0$, and apply the hypothesis of the old definition with $\frac{\varepsilon}{2}$. $\endgroup$
    – K. Jiang
    Commented Mar 14, 2023 at 3:58
  • $\begingroup$ @K.Jiang your first comment has errors, since if the original definition (which means using $< \delta$) is satisfied then it doesn't let you use $\leq \delta$ for the same $\varepsilon$. Instead, if the definition using $\leq \delta$ (the "alternate definition") is satisfied then so is the one using $< \delta$ (the "original definition"), while if the definition using $< \delta$ is satisfied then so is the one using $\leq \delta/2$, so you could rename $\delta/2$ as $\delta$. $\endgroup$
    – KCd
    Commented Mar 14, 2023 at 4:20
  • $\begingroup$ @KCd Thanks, I overlooked that. $\endgroup$
    – K. Jiang
    Commented Mar 14, 2023 at 4:23
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    $\begingroup$ You can use either, the important part is the $\epsilon>0$ and $\delta>0$. $\endgroup$
    – copper.hat
    Commented Mar 14, 2023 at 4:43
  • $\begingroup$ For the second question, the duplicate is math.stackexchange.com/q/1438766/139123. (Note that one of the guidelines here is to ask only one question per "question"), $\endgroup$
    – David K
    Commented Mar 14, 2023 at 6:09

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