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I got this question for homework, and I am not sure I fully understand it:

Prove that the next two statements are equivalent: $$\exists L\forall\epsilon\gt0\exists P:P\lt x\rightarrow\|f(x)-L\|\lt \epsilon$$ $$\exists L\forall\epsilon\gt0\exists P\gt0:P\lt x\rightarrow\|f(x)-L\|\lt \epsilon$$

Both statements say that $f(x)$ has a limit $L$ when $x\rightarrow\infty$. The definition of limit in infinity does not require anything from $P$, so what am I supposed to show here? Thanks

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  • $\begingroup$ The first statement makes no sense. What does "there exists greater than..." mean?? $\endgroup$ – Adam Rubinson May 6 '12 at 11:55
  • $\begingroup$ fixed it, not its ok $\endgroup$ – yotamoo May 6 '12 at 12:03
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    $\begingroup$ It should be clear that (2)$\implies$(1). Other implication: If you know that a $P$ with such properties exists, can you see that there exists also some $P'$ which has the same properties and it is, in addition to it, positive. $\endgroup$ – Martin Sleziak May 6 '12 at 12:11
  • $\begingroup$ Well "The definition of limit in infinity does not require anything from P, so what am I supposed to show here?" is basically the informal answer to the question. Statement 2 clearly implies Statement 1. The other way round is harder. Let's have athink... Edit: I like Martin's way for (1) implies (2) $\endgroup$ – Adam Rubinson May 6 '12 at 12:24
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You are supposed to formally prove that both statements imply each other.

Let $H(P)$ be "$P<x → ||f(x)-L||<ε$".

The second one implies the first because $∃P>0~H(P)$ implies $∃P~H(P)$.

The first one implies the second one because if you have $H(P)$ then you can find $P'$ such that $P'>0$ and $H(P')$. For example, $P'=|P|+1>0$ works.

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