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What if the epsilon-delta definition of a limit reversed the wording for $δ$ and $ϵ$: “for all $δ>0$, there exists an $ϵ>0$ such that, if $0<|x-a|<δ$, then $|f(x)-L|<ϵ$.” Would this definition still capture the concept of limit? Why or why not? If not, give a counter-example.

I want to say "no" this doesn't capture a limit because the definition of a limit must hold true for each and every epsilon. However, I'm not quite sure how to state this or what might be a counter-example.

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    $\begingroup$ Any bounded function $f$ would satisfy this property (just take $\varepsilon:=|L|+\sup|f|+1$, independently of $\delta$), no matter how "irregular", so continuity is not captured. $\endgroup$ – Milly Oct 27 '14 at 19:14
  • $\begingroup$ See this answer: math.stackexchange.com/a/1324644/72031 $\endgroup$ – Paramanand Singh Jun 5 '16 at 3:26
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Consider $f(x)=\frac{1}{x}$ defined for $x>0$. Then, $\lim_{x\to 0.5}f(x)=2$ in the usual sense. Yet, in your case, with $\delta=\frac{1}{2}$, there is no $\epsilon>0$ such that $0<|x-0.5|<\delta$ implies $|f(x)-2|<\epsilon$.

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