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Test Review 5 - limit of a constant sequence

  1. Oct 15, 2005 #1
    I need to prove that the limit of a constant sequence converges, using the definition of a limit.

    This is what I have:

    Let e > 0 be given.
    Then |sn - s| < e
    But sn = s for all sn, thus
    |s - s| < e
    |0| < e
    0 < e
    Thus N can be any number?

    This proof is simple, but I am making it complicated! Please help!
     
  2. jcsd
  3. Oct 15, 2005 #2
    Your solution is close but not correct. Remember, the definition of a limit involves [itex]\epsilon[/itex] AND [itex]\delta[/itex]. There is no mention of a [itex]\delta[/itex] in your proof.
     
  4. Oct 15, 2005 #3

    HallsofIvy

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    Staff Emeritus
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    No, Oxymoron, there is NO [itex]\delta[/itex] in the proof of a limit of sequenc! That's only for limit of a function in which the variable is takes on continuous values.

    CMurphy, your proof is completely correct: N can be taken to be anything. It really is that easy!
     
  5. Oct 15, 2005 #4
    Sorry guys, didn't realize it was a sequence. Halls is 100% correct, and so are you colleen. However, if it was a function then you need the delta. I hope you realize that you didnt need the delta because you only have a sequence.
     
  6. Oct 16, 2005 #5
    That's a truth "with modification", since the epsilon-delta definition of limits (or continuity) can be formulated using only sequences.
     
    Last edited: Oct 16, 2005
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