1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of bounded functions

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Let f and g be real-valued functions defined on A ⊆ R and let c ∈ R be a cluster point of A. Suppose that f is bounded on a neighborhood of c and that limx→c g(x) = 0. Prove that limx→c f(x)g(x) = 0.

    2. Relevant equations

    3. The attempt at a solution
    This isn't a very hard question, but it has to be done with no assumptions from calculus (Analysis 1).
    Is it sufficient to say:
    since f is bounded by (c-δ, c+δ) for some δ>0,
    then limx→cf(x) = L is bounded by (f(c-δ), f(c+δ)),
    and since f(x0) is a real number, for any x0 in that interval,

    limx→c f(x)g(x) = limx→c f(x)limx→c g(x) = L * 0 = 0

    I'm just not too sure what would be a formal proof...
     
  2. jcsd
  3. Oct 18, 2011 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    What do you mean with "if f is bounded by (a, b), then the limit is bounded by (f(a), f(b))"? Do you mean: if |f(x)| < a then |lim f(x)| < f(a)?
    Because that is not necessarily true, if it even makes sense.

    I think that the formal proof you are after uses the epsilon-delta definition, i.e.
    [tex]\forall_{\epsilon > 0} \exists_{\delta(\epsilon) > 0} : |x - c| < \delta \implies |f(x)g(x)| < \epsilon[/tex]
    Of course you already know that
    [tex]\forall_{\epsilon > 0} \exists_{\delta'(\epsilon) > 0} : |x - c| < \delta' \implies |g(x)| < \epsilon[/tex]
     
  4. Oct 18, 2011 #3
    Thank you, that makes sense
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook