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Math Help - Limit of Maximum

  1. #1
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    Limit of Maximum

    Anyone can help me ?

    QUESTION :
    for all x in (0,1), n is Natural number

    f_n (x) := max{1/x , n }

    find lim f_n (x) when n -> infinity.
    Last edited by younhock; November 16th 2011 at 06:29 AM.
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  2. #2
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    Re: Limit of Maximum

    Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?

    If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.
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  3. #3
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    Re: Limit of Maximum

    Quote Originally Posted by HallsofIvy View Post
    Do you mean {1/n, n}? If so then you are looking at "max(1/2, 2)", "max(1/3, 3)", "max(1/4, 4)"...:"max{1/1000000, 1000000}", ... What are those numbers? What is the limit as n goes to infinity?

    If you really mean {1/x, n} for fixed x it is essentially the same: "max(1/x, 2)", "max(1/x, 3)", "max(1/x 4)"...:"max{1/x 1000000}", where the first number is some fixed finite number.
    i had updated the question to make it clear.
    I was interested to know what is the limit function which f_n approaching to .
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  4. #4
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    Re: Limit of Maximum

    It's pretty straight forward if you write out a few terms in the sequence of functions.
    For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So f_n(x)= 1/x for x< 1/n and f_n(x)= n for x> 1/n, In particular, f_1(x)= 1/x for all x in (0, 1). f_2(x)= 1/x for 0< x<= 1/2, f_2(x)= 2 for 1/2<= x< 1. f_3(x)= 1/x for 0< x<= 1/3 and f_3(x)= 3 for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, f_n(x)> N.
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  5. #5
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    Re: Limit of Maximum

    Quote Originally Posted by HallsofIvy View Post
    It's pretty straight forward if you write out a few terms in the sequence of functions.
    For x in (0, 1) (in particular for x positive) 1/x> n when x< 1/n. So f_n(x)= 1/x for x< 1/n and f_n(x)= n for x> 1/n, In particular, f_1(x)= 1/x for all x in (0, 1). f_2(x)= 1/x for 0< x<= 1/2, f_2(x)= 2 for 1/2<= x< 1. f_3(x)= 1/x for 0< x<= 1/3 and f_3(x)= 3 for 1/3<= x< 1. Do see what is happening? Given any x, there is eventually an integer N> 1/x so for n> N, f_n(x)> N.
    i see, so we will get f_n(x) -> infinity eventually. Thanks yea.
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