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Confused by a simple limit

  1. Dec 7, 2011 #1
    What is the value of
    [itex]\lim\limits_{n\rightarrow\infty}\sup \left\{\frac{n}{x^n}:x\in\left( 1; \infty\right)\right\}[/itex]
    It seems to be 0, but what if [itex]x = 1+\frac{1}{n}[/itex]? In that case [itex]x^n = e[/itex] and the above limit is then [itex]+\infty[/itex], isn't it? I have a feeling I'm somehow wrong, but if I'm not, for what x is the above limit equal to 0 ?
     
  2. jcsd
  3. Dec 7, 2011 #2
    function goes to 0 when x>1

    thats the answer.
     
  4. Dec 7, 2011 #3
    Sorry, my first post was wrong, this is the edit. I think that if you do that, then x = 1, and is no longer satisfying the bounds of being between 1 and infinity.
     
  5. Dec 7, 2011 #4
    @EternityMech, would you elaborate? Is [itex]1+\frac{1}{n} = 1[/itex]? I'm sure it isn't because raising both sides to the power of n gives [itex]e = 1[/itex]
     
  6. Dec 7, 2011 #5
    Yes I realized my error and changed my post.

    EDIT: ooh and it looks like you realized that i realized my error and you also changed your post. well...thats good that this is all sorted out.
     
  7. Dec 8, 2011 #6
    its a lim sup of sequence of numbers or sequence of sets ? if numbers show me what are a1,a2,a3,.....and if sets show what are X1,X2,X3,......
     
  8. Dec 8, 2011 #7
    First of all, x is fixed. You can not replace it by a terms that depends on n, if you do, you change the sequence. What the sequence is completely determined by how n appears in it.

    This is limsup so first you have to find supremum over m>n and then take the limit. I am assuming you are defining the supremum as follows (it may change but the general convention is this)

    [itex]\sup_{m>n} \left\{\frac{m}{x^m}:x\in\left( 1; \infty\right)\right\}[/itex]

    However in our case the terms approach to zero (divide n+1th element by the nth element and take the limit, you will see that it is less than 1 so using theorems from calculus you can say that the limit exists and is zero). So when your sequence has a limit, liminf and limsup are equal to it and to each other (for instance on the contrary consider a sequence of points which converge to points on a sine graph, they do not have a limit but their limsup and liminf is 1 and -1). Thus the answer is zero.

    More explicitly you can show that after a certain value of n (when we pass over the maximum of [itex]\frac{m}{x^m}[/itex]) then
    [itex]sup_{m>n} \frac{m}{x^m} = \frac{n+1}{x^{n+1}} [/itex] and so taking the limit gives you the result.
     
    Last edited: Dec 8, 2011
  9. Dec 8, 2011 #8
    I guess I should add some context. I'm given [itex]f_n(x)=\frac{n}{x^n}[/itex]. I need to analyze the convergence of this sequence of functions for all [itex]x > 1[/itex]. First I see that [itex]f(x) = \lim\limits_{n\rightarrow\infty} \frac{n}{x^n} = 0[/itex]. I then need to check whether [itex]\forall\epsilon>0, \exists N,\forall n > N, \forall x: |f-f_n|(x) < \epsilon[/itex] to know if the convergence is uniform. I do that using the expression in my first post. I'm sure that [itex]x[/itex] can be a function of [itex]n[/itex] because [itex]\forall x[/itex] goes after [itex]\exists N[/itex].

    @vrmuth, I'm not sure how to define it. I guess it's a set of functions of n. Either way, the set is not discrete.
     
  10. Dec 8, 2011 #9
    well you should definitely distinguish between sequences of functions and sequences of points :) your first post looks more like a limit of sequence of points rather than functions. Still it is a bit strange to define x interms of n since it is just a free parameter no?

    The limit fn(1+n) as n goes to infinity is not a point-wise limit but you just select the value of each fn at the point 1+n and look whether if that has a limit. I can't see what that would have to do with this question since we want to look at pointwise limits or uniform limit.
     
    Last edited: Dec 8, 2011
  11. Dec 8, 2011 #10
    What I can't distinguish between is functions, numbers and sets. A function because it has an argument n, a number if n is a constant and a set if n is any natural number.

    If limit fn(1+1/n) is not a pointwise limit, I assume that this sequence does not converge uniformly. In that case, my question is, for what x does it converge uniformly?
     
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