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I am going through Spivak's Calculus right now and I am trying the exercises in Chapter 5. Spivak gives the following definition of the limit $L$ of a function as $x$ approaches $a$:

$$\forall \epsilon>0, \exists\delta>0:0<\left\lvert x-a \right\rvert<\delta \implies \left\lvert f(x)-L \right\rvert<\epsilon.$$

To prove that a limit doesn't exist, acoording to Spivak, one must use the following negation:

$$\forall L, \exists\epsilon>0:\forall \delta>0, \exists x: 0<\left\lvert x-a \right\rvert<\delta \ \text{and} \left\lvert f(x)-L \right\rvert\geqslant\epsilon.$$

Now my question is, since $\delta$ can depend on $\epsilon$ when proving that a limit exists, why haven't I seen anyone pick $\epsilon$ based on $\delta$ when doing the opposite? What I have noticed in those cases is that the author would pick an abitrary number $\beta$ in place of $\delta$, find a suitable $\epsilon$ in terms of that number and then pick $0<\left\lvert x-a \right\rvert < \min(\beta, \delta)$ to obtain the desired contradiction for any $\delta>0$

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    $\begingroup$ Wrong negation. $\endgroup$ Commented Jun 20 at 10:22
  • $\begingroup$ $\neg(p\to q)=\neg(\neg p\lor q)=p\land\neg q$ $\endgroup$ Commented Jun 20 at 10:26
  • $\begingroup$ Thank you! I corrected the mistaken negation. $\endgroup$ Commented Jun 20 at 10:27
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    $\begingroup$ Of course you choose $\delta$ according to the choice of $\epsilon$ in showing that a limit exists. How else would you do it? $\endgroup$ Commented Jun 20 at 10:38
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    $\begingroup$ When you have a $\forall$ followed by an $\exists$, the $\exists$ variable depends on the $\forall$ variable. So in the first case $\delta$ depends on $\epsilon$, whereas in the second $\epsilon$ depends on $L$ while $x$ depends on $\delta$ (and also on $L$ and $\epsilon$, since those are used to "set up" the statement about all $\delta$). $\endgroup$ Commented Jun 20 at 13:21

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When you have a statement with quantifiers, like $(\forall \epsilon>0) (\exists \delta>0) \ldots,$ the existentially quantified variables (e.g. $\exists \delta$) can depend on universally quantified variables (e.g. $\forall \epsilon$) that precedes them. So $\delta$ can depend on $\epsilon$ here, but not reversely.

For $(\forall L)(\exists\epsilon>0)(\forall\delta>0)\ldots$ the above rule gives that $\epsilon$ can depend on $L$ but not on $\delta.$

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When you use, $$\forall L\in\mathbb R,\exists\varepsilon>0\text{ s.t. }\forall\delta>0,\exists x\in\mathbb R\text{ s.t. }0<|x-a|<\delta \text{ and }|f(x)-L|\ge \varepsilon$$ what's happening is, given your arbitrary $L\in\mathbb R$, the statement guarantees the existence of a $\varepsilon>0$ such that the rest of the statement holds. This $\varepsilon$ should only depend on $L$, since we only told it our arbitrary $L$.

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