Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differentiable function, limits, sequence
  1. Limit of sequence (Replies: 1)

  2. Limits of a sequence (Replies: 1)