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Differentiable function, limits, sequence

  1. Feb 6, 2008 #1
    f is differentiable on [tex](a,\infty)[/tex] and


    I am trying to prove that there exists a sequence [tex]\{x_n\}, x_n\rightarrow \infty,[/tex] such that [tex]f'(x_n)\rightarrow A.[/tex]

    Any help would be appreciated.
  2. jcsd
  3. Feb 6, 2008 #2


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    Can you give a heuristic explanation why it should be true?
  4. Feb 7, 2008 #3


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    then what is the limit of [f(x) - f(a)]/[x-a], for any a?
  5. Feb 7, 2008 #4
    By MVT, it would equal f'(c), where c is in (a,x). But I don't see how I can get a sequence from this fact.

    As x gets larger, there is always a c_n in (a,x) such that f'(c)=[f(x) - f(a)]/[x-a]. Perhaps it can be said that the sequence of x_n approaches A, but I don't know how to get there.
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