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Showing a limit exists using differentiability

  1. Feb 11, 2016 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    Assume f:(a,b)→ℝ is differentiable on (a,b) and that |f'(x)| < 1 for all x in (a,b). Let an
    be a sequence in (a,b) so that an→a. Show that the limit as n goes to infinity of f(an) exists.
    2. Relevant equations
    We've learned about the mean value theorem, and all of that fun stuff.

    3. The attempt at a solution
    I don't really know where to start so I brainstormed a couple of things I noticed

    I know that since |f'(x)| is always less than one, any sequence of points will be bounded. Since they are bounded they are cauchy.
    I also know that
    ## lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} ## exists for all x. I assume this should also mean that ## lim_{n \rightarrow \infty} \frac{f(a_n)-f(a)}{a_n - a} ## exists. I think with these two facts I can construct a proof but I don't know if either of the two are correct
     
  2. jcsd
  3. Feb 11, 2016 #2

    Samy_A

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    Homework Helper

    Hint:

    Take two element of the sequence, ##a_n## and ##a_m##, and apply the mean value theorem for these two values.
    That should give you an interesting upper bound for ## |f(a_n)-f(a_m)|##.

    It is not true that a bounded sequence is Cauchy.
    But, a convergent sequence is a Cauchy sequence, and in ℝ a Cauchy sequence has a limit. This, together with the upper bound for ## |f(a_n)-f(a_m)|## should lead to a solution.
     
    Last edited: Feb 11, 2016
  4. Feb 11, 2016 #3

    B3NR4Y

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    Ok so applying the mean value theorem, there is an x in (am,an) so that ## \frac{f(a_m) - f(a_n)}{a_m - a_n} = f'(x) ## taking the absolute value of both sides, ## |f(a_m) - f(a_n)| < |a_m - a_n| ##. Since an is convergent, |am - an| is always less than some epsilon greater than zero (because it is a convergent sequence in R and therefore Cauchy), and so |f(am) - f(an)| is also always less than that epsilon. This is the definition of a cauchy sequence, so the sequence f(an) must have a limit.

    Is this the right path?
     
  5. Feb 11, 2016 #4

    Samy_A

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    Yes.
    Essentially you have that ##(f(a_n))_n## is a Cauchy sequence, therefore convergent.

    Maybe a remark: where you say "is always less than some epsilon", you should add "for n and m sufficiently large". But that is probably what you implicitely meant.
     
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