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For my analysis homework I am asked to prove or disprove that if you switch the quantifiers of the definition of a limit, then the definitions are equivalent. More specifically we are asked to prove or disprove for $\lim_{x\to c}f(x)=L$ that these definitions are equivalent,

$\forall \epsilon>0\; \exists\delta>0\; s.t. \forall x\in A\; \text{if}\; |x-c|<\delta, \text{then}\; |f(x)-L|<\epsilon $

$\exists\delta>0\; \forall \epsilon>0 s.t. \forall x\in A\; \text{if}\; |x-c|<\delta, \text{then}\; |f(x)-L|<\epsilon $

My idea is that they are not equivalent. My reasoning is that in the second definition there is an existence claim for delta followed by for all epsilon. This would imply that epsilon depends on delta which would make every limit exist because there is a delta for every epsilon meaning $|x-c|<\delta$ would always imply $|f(x)-L|<\epsilon$.

I am wondering if this is correct reasoning or if I am missing something. Thank you!

Eric Wofsey
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yastown
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2 Answers2

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I cannot agree or disagree with you, since I fail to see why is it that you say that “This would imply that epsilon depends on delta”.

Anyway, the second assertion implies that $f(x)$ is constant and equal to $L$ on $(c-\delta,c+\delta)$ (since, on that interval, $|f(x)-L|$ is smaller than every number greater than $0$), and therefore, if, say, $f(x)=x$, then the first assertion holds (with $c=L=0$), but not the second one.

José Carlos Santos
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  • My reasons for saying that epsilon depends on delta is because we have that there exists delta for every epsilon. So for any epsilon we are given we can find a delta. – yastown Apr 07 '21 at 21:33
  • But what the second definition tells us is that there is some $\delta>0$ such that a certain thing occurs for every $\varepsilon>0$. If it is for every $\varepsilon>0$, it makes no sense to say that $\varepsilon$ depends upon $\delta$. – José Carlos Santos Apr 07 '21 at 21:36
  • Gotcha, thank you! – yastown Apr 07 '21 at 21:42
  • However, at this point I am confused on how to prove that they are not equivalent. – yastown Apr 07 '21 at 21:43
  • What's the problem with my answer? – José Carlos Santos Apr 07 '21 at 21:46
  • I am a bit confused on how you deduced that f(x) is constant and equal to L on the interval (−,+). – yastown Apr 07 '21 at 21:52
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    Suppose that there is a $x\in(c-\delta,c+\delta)$ such that $f(x)\ne L$. Then $|f(x)-L|>0$. So, take $\varepsilon=|f(x)-L|$, and it will be true that $|f(x)-L|\geqslant\varepsilon$. This is impossible, since we are assuming that, for every $x\in(c-\delta,c+\delta)$ and every $\varepsilon>0$, $|f(x)-L|<\varepsilon$. – José Carlos Santos Apr 07 '21 at 21:57
  • I see, that clears up my question. Thank you! – yastown Apr 07 '21 at 21:59
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The second definition would be a non sense.

Let us choose a $\delta$ among the existing ones. Then

$$m:=\max_{|x-c|<\delta}|f(x)-L|$$ is a positive number (unless $f(x)$ is identically the constant $L$), and it is false that $\forall \epsilon>0:m<\epsilon$.