Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
The first statement makes no sense. What does "there exists greater than..." mean?? –  Adam Rubinson May 6 '12 at 11:55
    
fixed it, not its ok –  yotamoo May 6 '12 at 12:03
1  
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. –  Martin Sleziak May 6 '12 at 12:11
    
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) –  Adam Rubinson May 6 '12 at 12:24
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.