Results 1 to 3 of 3

Thread: Proofs using ε

  1. #1
    Junior Member
    Joined
    Oct 2009
    From
    Aberystwyth, Wales
    Posts
    36

    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:
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    Quote Originally Posted by renlok View Post
    why not use
    $\displaystyle \left|{x_n - l}\right| = 0$?
    Because $\displaystyle \left| {\frac{1}{n} - 0} \right| \ne 0,~~\forall n$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proofs
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Mar 2nd 2010, 03:54 AM
  2. lim sup and lim inf proofs
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: Feb 24th 2010, 07:02 PM
  3. More Proofs
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Feb 13th 2008, 07:05 PM
  4. Proofs
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 3rd 2008, 04:23 AM
  5. Replies: 3
    Last Post: Oct 6th 2007, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum