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Limit of x^n where 0<x<1

  1. Oct 17, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Almost like the title says : Let [tex]f_n (x)=x^n[/tex] for [tex]0 \leq x \leq 1[/tex]. Find the pointwise limit of [tex]f_n[/tex] and show that the convergence is not uniform.
    What happens if [tex]x \in [0,1)[/tex]?





    2. The attempt at a solution
    [tex]\lim _{n \to \infty} f_n = \lim _{n \to \infty} x^n = 0[/tex] if [tex]x[/tex] is strictly lesser than [tex]1[/tex] and it's worth [tex]1[/tex] if [tex]x[/tex] equals [tex]1[/tex]. (I wonder if I have to prove it via the definition of limits).
    I don't have much ideas about the non uniform convergence. Checking up some wikipedia, I think that if I can show that if [tex]a_n[/tex] does not tend to [tex]0[/tex] when [tex]n[/tex] tends to [tex]\infty[/tex], I'm done. Where [tex]a_n=sup |f_n (x)-f(x)|[/tex].
    I get that [tex]a_n = sup |x(x^{n-1}-1)|[/tex] but I think it's worth [tex]0[/tex], hence implying the uniform convergence of [tex]f_n[/tex]...


    Answering the last question : By intuition the convergence is uniform.
     
  2. jcsd
  3. Oct 17, 2009 #2
    The convergence is pointwise because the convergence of the sequence depends on what points of x you choose. Choosing x in [0,1), x=1, and x > 1 gives you a different limit.
    You can use the definition of uniform convergence here: http://en.wikipedia.org/wiki/Uniform_convergence#Definition. If you are having trouble with pointwise vs. uniform convergence, compare the position of the phrase 'for all x in S' in the two definitions.
     
  4. Oct 17, 2009 #3

    fluidistic

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    Gold Member

    Ok thanks, I'll try my best.
     
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