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Limits at infinity, lim xF(x) = L then lim (f(x)=0

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    show that if F:(a,∞) -->R is such that lim xF(x) = L, x --> ∞, where L is in R, then lim F(x) = 0, x --> ∞.

    2. Relevant equations

    3. The attempt at a solution

    Let F:(a,∞) →R is such that lim xF(x) = L, x → infinity, where L is in R. Then there exists an α> 0 where given ε, there exist k(ε) for all x > k then ε > max{1 , ([L]+1)/x} Therefore [xF(x) - L] < 1 whenever x > α. Therefore [F(x)] < ([L]+1)/x. Thus [F(x)-0] < ε Then lim F(x) =0 as x → ∞. This is what I have but it doesn't look right to me.
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 15, 2012 #2
    Proof: let f:(a,∞)→ℝ such that lim xf(x)=L where L in ℝ. Since lim xf(x) = L, there exists α>0 where |xf(x)-L| < 1 for all x > α. Therefore |f(x)|<(|L|+1)/x for x >α. Pick ε = m where there exist δ-neighborhood Vδ(c) of c and x is in A π Vδ(c), there exists m>0, m = |L|+1 then |f(x)| < M, for all X, therefore |f(x)-o|<M=
    thus the limitx→∞ f(x) =0.

    Does this proof make more sense? Am i still missing something?

  4. Mar 15, 2012 #3


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    You have some undefined quantities.

    You can't simply pick the ε in the part where you prove that lim x→∞ f(x) = 0 .

    It may help for you to state, in ε - M language, what it means that lim x→∞ f(x) = 0 .
  5. Mar 15, 2012 #4
    This can be proved in one line.

    Lt x->infinity
    xf(x) =L (where L is finite)

    So Lt x->infinity f(x) = L/x (How?)

    What do you see??
  6. Mar 15, 2012 #5
    I think i see L/x going to zero as x goes to infinity since L is finite.
  7. Mar 15, 2012 #6
    Correct :-)
  8. Mar 15, 2012 #7
    just curious, what is meant by ε - M language?
  9. Mar 16, 2012 #8


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    Given an ε>0, there exists an integer, M, such for all x > M, ...
  10. Mar 16, 2012 #9
    Thanks... I am an idiot!
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