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Math Help - pointwise limit.

  1. #1
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    pointwise limit.

    if Fn(x) = x^(1/n) with Fn: [0,infinity) -> R

    Find the pointwise limit of Fn(x)

    this is what i have done:

    x=0 Fn(x) -> 0
    x!=0 Fn(x) -> 1

    Is this correct... Im new to this so im not quite show about my answer

    please Help
    Last edited by mr fantastic; November 10th 2010 at 02:14 PM. Reason: Title.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dreamer78692 View Post
    if Fn(x) = x^(1/n) with Fn: [0,infinity) -> R

    Find the pointwise limit of Fn(x)

    this is what i have done:

    x=0 Fn(x) -> 0
    x!=0 Fn(x) -> 1

    Is this correct... Im new to this so im not quite show about my answer

    please Help
    Ok, so how do you think we should show it (you're answer is correct)
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  3. #3
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    Do you mean how I arrived at my answer???
    If so I used a theorem that states that a(1/n) -> 1 as n -> infinity where a is an element of R.

    Sorry there is a second part to this question that is:

    Does Fn converge uniformly on [0,1]

    My answer is no since sup|x(1/n)| -> 1 as n->infinity

    AND

    Does Fn converge uniformly on [1/2,1]

    My answer is the same as above

    I seriously doubt this is correct

    Could you please help with this one
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dreamer78692 View Post
    Do you mean how I arrived at my answer???
    If so I used a theorem that states that a(1/n) -> 1 as n -> infinity where a is an element of R.

    Sorry there is a second part to this question that is:

    Does Fn converge uniformly on [0,1]

    My answer is no since sup|x(1/n)| -> 1 as n->infinity

    AND

    Does Fn converge uniformly on [1/2,1]

    My answer is the same as above

    I seriously doubt this is correct

    Could you please help with this one
    Well, it's surely not uniformly convergent since each of the f_n(x)'s is continuous but the limit isn't (this is on [0,\infty)) That said, I would try re-evaluating the second question.
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