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Math Help - Proofs using ε

  1. #1
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    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 \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
    \left|{x_n - l}\right| = 0?

    analysis just makes no sense D:
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  2. #2
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    Quote Originally Posted by renlok View Post
    why not use
    \left|{x_n - l}\right| = 0?
    Because \left| {\frac{1}{n} - 0} \right| \ne 0,~~\forall n.
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  3. #3
    Member mohammadfawaz's Avatar
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    In fact, all definitions related to limits contain these \epsilon notations. The idea here is that for all \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 x_n and the limit is less than that \epsilon. In other words, the sequence is approaching the limit.
    In fact, you can't just say \left|{x_n - l}\right| = 0 because you may never find a value of x_n that is equal to l. For example, if x_n = \frac{1}{n}, then we claim that the limit of x_n is zero as n goes to infinity. However, you can't give any number n such that \frac{1}{n} = 0!. In fact, for any given \epsilon, look at |\frac{1}{n}-0|=\frac{1}{n} which can be made less than \epsilon if you choose sufficiently large n, namely n>N = \frac{1}{\epsilon}. This proves that the limit is zero! That is, for any given \epsilon>0, you can find a natural N such that if n>N, then |\frac{1}{n}-0|<\epsilon
    hope this helps!
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