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Limit Theorems

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Theorem: Let f:[a, ∞)→ R. The following are equivalent.

    i) lim ƒ(x) = A as x→∞
    ii) For all sequences {xn
    in [a,∞) with lim xn = ∞

    we have lim f(xn) = A.

    2. Relevant equations

    For any ε > 0, |ƒ(x)-A| < ε if x < N

    3. The attempt at a solution

    I probably have this wrong, but I think I should show that for any N > 0 in [a, ∞) there exists an xn0 > N if n≥ n0

    I imagine that to tie this in to the idea of continuity, I'd have to come up with an arbitrary function f(c) to get |f(xn)-f(c)| < ε when x > N

    I just don't know how to say f(x) has the same domain as f(xn) without just stating it.
     
  2. jcsd
  3. Feb 5, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    The second inequality is backwards.

    [itex]f(x) \to A[/itex] as [itex]x \to \infty[/itex] if and only if for all [itex]\epsilon > 0[/itex] there exists [itex]R > 0[/itex] such that for all [itex]x \in [a, \infty)[/itex], if [itex]x > R[/itex] then [itex]|f(x) - A| < \epsilon[/itex].

    [itex]x_n \to \infty[/itex] if and only if for all [itex]R > 0[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex], if [itex]n \geq N[/itex] then [itex]x_n > R[/itex].

    That, sadly, is wrong. It looks like a conflation of the definitions of "[itex]x_n \to \infty[/itex]" (which I have given above) and "the sequence [itex](x_n)[/itex] is not bounded above" (which is "for all [itex]R > 0[/itex] there exists [itex]n \in \mathbb{N}[/itex] such that [itex]x_n > R[/itex]"). Clearly if [itex]x_n \to \infty[/itex], as we are assuming, then it must follow that [itex](x_n)[/itex] is not bounded above.

    This exercise has nothing to do with continuity. A function is continuous at [itex]c[/itex] if and only if [itex]\lim_{x \to c} f(x) = f(c)[/itex]. But this exercise is concerned with the limit as [itex]x \to \infty[/itex], and since [itex]\infty[/itex] is not a real number there is no such thing as [itex]f(\infty)[/itex]. Instead we are given [itex]\lim_{x \to \infty} f(x) = A[/itex].

    To show that (i) implies (ii), you must assume that [itex]f(x) \to A[/itex] as [itex]x \to \infty[/itex], then take an arbitrary [itex]x_n \to \infty[/itex] and show that for all [itex]\epsilon > 0[/itex] there exists an [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex], if [itex]n \geq N[/itex] then [itex]|f(x_n) - A| < \epsilon[/itex].

    Proving that (ii) implies (i) is slightly harder.
     
  4. Feb 5, 2014 #3
    Thanks.
     
    Last edited: Feb 5, 2014
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