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Thread: Proofs using ε

  1. #1
    Junior Member
    Oct 2009
    Aberystwyth, Wales

    Proofs using ε

    I'm trying to do some revision for my analysis exam and when it comes to proofs where ε is used I dont understand what these proofs mean.

    Like for example
    Given $\displaystyle \epsilon > 0, \forall n \in N: \left|{x_n - l}\right| < \epsilon$
    which is something to do with the limit of the sequence xn being l

    why not use
    $\displaystyle \left|{x_n - l}\right| = 0$?

    analysis just makes no sense D:
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  2. #2
    MHF Contributor

    Aug 2006
    Quote Originally Posted by renlok View Post
    why not use
    $\displaystyle \left|{x_n - l}\right| = 0$?
    Because $\displaystyle \left| {\frac{1}{n} - 0} \right| \ne 0,~~\forall n$.
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  3. #3
    Member mohammadfawaz's Avatar
    Feb 2010
    Lebanon - Beirut
    In fact, all definitions related to limits contain these $\displaystyle \epsilon$ notations. The idea here is that for all $\displaystyle \epsilon>0$, you can still find values of n larger than some N>0 (this should be added to your statement: Not for all n!)and where the difference between $\displaystyle x_n$ and the limit is less than that $\displaystyle \epsilon$. In other words, the sequence is approaching the limit.
    In fact, you can't just say $\displaystyle \left|{x_n - l}\right| = 0$ because you may never find a value of $\displaystyle x_n$ that is equal to l. For example, if $\displaystyle x_n = \frac{1}{n}$, then we claim that the limit of $\displaystyle x_n$ is zero as n goes to infinity. However, you can't give any number n such that $\displaystyle \frac{1}{n} = 0!$. In fact, for any given $\displaystyle \epsilon$, look at $\displaystyle |\frac{1}{n}-0|=\frac{1}{n}$ which can be made less than $\displaystyle \epsilon$ if you choose sufficiently large n, namely $\displaystyle n>N = \frac{1}{\epsilon}$. This proves that the limit is zero! That is, for any given $\displaystyle \epsilon>0$, you can find a natural N such that if $\displaystyle n>N$, then $\displaystyle |\frac{1}{n}-0|<\epsilon$
    hope this helps!
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