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Simple analysis proof: x^n -> a^n as x -> a

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that x^n approaches a^n as x approaches a.


    2. Relevant equations



    3. The attempt at a solution
    I understand the concept here ... need to find a delta>0 for epsilon>0 s.t. |x-a|<delta implies |x^n-a^n|<epsilon. For some reason I can't solve this one. Thanks.
     
  2. jcsd
  3. Sep 29, 2009 #2

    Dick

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    You know you can factor x^n-a^n=(x-a)*(x^(n-1)+x^(n-2)*a+ ... + a^(n-1)), right?
     
  4. Sep 29, 2009 #3
    Oh! Somehow I missed that. Thank you very much for your help.
     
  5. Sep 29, 2009 #4
    Dick assumes n is a positive integer, which you did not tell us under "all variables and given/known data" ... so are we to assume n is NOT know to be a positive integer?
     
  6. Oct 2, 2009 #5
    still, how do you show that (x^(n-1)+x^(n-2)*a+ ... + a^(n-1)) is < epsilon?
     
  7. Oct 2, 2009 #6

    HallsofIvy

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    You don't, and no one has claimed it is. You want [itex]|x^n- a^n|= |x- a||x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|< \epsilon[/itex] for |x- a| small enough. That is the same as choosing [itex]|x- a|< \epsilon/|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|[/itex] and you can do that by finding any positive lower bound on [itex]|x^{n-1}+ ax^{n-2}+ ...+ a^{n-1}|[/itex] for x close to a.
     
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