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Absolute value of a limit

  1. Feb 6, 2009 #1
    Hi all,
    I'm wondering about this question

    I can prove that if lim_{n->inf} (a) = L then lim_{n->inf}abs(a) = abs(L)
    however.. is the converse true?
    thx
     
  2. jcsd
  3. Feb 6, 2009 #2

    quasar987

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    Consider the map f:R-->R defined by

    f(x)= 1 if x is rational and =-1 if x is irrational

    Then |f| is identically 1 so it is everywhere continuous, while f is nowhere continuous by the density property of the rational and irrational numbers.

    In particular, for instance, let x_n be a sequence converging to 0 that contains an infinity of rational numbers and an infinity of irrational numbers and set a_n :=f(x_n). Then |a_n|-->1 but a_n does not converge because it has a subsequence converging to 1 and another converging to -1.

    However, if |a_n|-->L and a_n converges, then it must be that a_n-->±L because if a_n-->P, and if

    |a_n-P|<epsilon,

    as soon as n>N, then because of the triangle inequality

    ||a_n|-|P||<=|a_n-P|,

    it follows that

    ||a_n|-|P||<epsilon

    as soon as n>N also.

    But this is the same as saying that |a_n|-->|P|. So L=|P|. So P=±L.
     
  4. Feb 6, 2009 #3
    Considering the constant sequnce -1 would also do the job. Its absolute value convreges to 1, so you can choose L=1 (of course you could also choose l=-1:smile:, but you don't have to.)
    Then the sequnece does not converge to L...
     
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