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Proving limit of product of functions?

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be bounded on the interval c < x < infinity, and let lim g(x)x-->infinity = 0. Prove that lim f(x)g(x)x-->infinity = 0. Does this follow directly from Theorem 19?

    2. Relevant equations

    Theorem 19

    Assuming that f and g are each defined on a deleted neighborhood of x = b, and that lim f(x)x-->b = A and lim g(x)x-->b = B, then it is true that

    lim (f(x) + g(x))x-->b = A + B

    lim f(x)g(x)x-->b = AB

    lim f(x)/g(x)x-->b = A/B if B does not equal 0.

    3. The attempt at a solution

    It cannot follow from the Theorem because you can't be sure the limit of f(x) even exists. So, do you have to prove that lim f(x) exists and then it would follow from the theorem that lim f(x)g(x) = A*0 = 0?
     
  2. jcsd
  3. Oct 20, 2009 #2

    Dick

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    You are right. It doesn't follow directly from Theorem 19. f(x) doesn't need to have a limit. But f(x) being bounded is enough to show lim x->infinity f(x)g(x) is zero. It's a squeeze argument, right?
     
  4. Oct 20, 2009 #3
    Since f is bounded on the interval (c, infinity), f(x)g(x) will be bounded on the same interval. How do you know g(x) is bounded on it too?

    Assume f(x) <or= g(x) <or= f(x)g(x). Since lim g(x) = 0, lim f(x) = lim f(x)g(x) = 0 by the Squeeze Theorem.
     
  5. Oct 20, 2009 #4

    Dick

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    Don't give me this gibberish. You don't need 'g(x) bounded'. f(x) is bounded. -M<=f(x)<=M where M is a bound for |f(x)|. g(x)->0. If you can't put that into a coherent argument using the Squeeze Theorem, I can't help you.
     
  6. Oct 20, 2009 #5
    Assume f(x) <or= g(x) <or= f(x)g(x). Since lim g(x) = 0, lim f(x) = lim f(x)g(x) = 0 by the Squeeze Theorem.

    Is that not correct logic then? I'm sorry if I seem to be processing this slow. I really am trying my best. I don't have a very strong background in Math.
     
  7. Oct 20, 2009 #6

    Dick

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    Ok, is it signs that are confusing you? If -M<=f(x)<=M (which is f bounded). Then 0<=|f(x)g(x)|<=M*|g(x)|. g(x)->0. Doesn't that show f(x)g(x)->0?
     
  8. Oct 20, 2009 #7
    Ohhhhhhhh! That is more simple than I thought it would be. I'm really sorry you had to hit me over the head with it for me to understand. But, I really do appreciate it. I know I am not very good at this.
     
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