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Math Help - May I know how to prove the following limit?

  1. #1
    Junior Member
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    Smile May I know how to prove the following limit?

    Dear All,

    I am looking for a proof of the following:

    For x>0, 0<f(x)<1, df(x)/dx<0, prove that Lim[x*f(x)*df(x)/dx]=0 as x approaches positive infinite.

    Thank you very much for the help!
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  2. #2
    Senior Member Dinkydoe's Avatar
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    You can show that for large n we have f(n)\approx 1/n^p+c for some c\in [0,1), p> 0
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  3. #3
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    Try to prove ...

    Proof:
    1. since df(x)/dx =0 for large x, therefore, df(x)/dx=-a/x^p, a>0, p>0
    2. integral above so that: f(x)=bx^(1-p)+c<1, b=-a/(1-p), 0<c<1,
    3. as 0<f(x)<1, we must have: (1-p)<0, or p>1
    4. Therefore: x*f(x)*df(x) < x*1*(-ax^(-p))=-a*x^(1-p)
    5. when p>1, or 1-p<0, -a*x^(1-p)=0
    Q.E.D

    I am not comfortable with the proof:
    For step 1, how do i know: df(x)/dx=-ax^(-p)? why not df(x)/dx=exp(-ax)? or any other function, and its compund functions (there are infinite of such functions), that equal 0 as x approaches infinite
    For step2, can I integrate the df(x) as this is approximate only
    Step,3,4,5 are ok to me, but not sure they are really reasonably rigorous

    Thank you very much!
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  4. #4
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    Maybe this is a better one...

    1. Suppose Lim ABS(x*df)>a, a>0.
    2. Since df<0 (which means always -df>0 ), x>0, we have: lim df<-a/x. Integral to have: f(x)<-a*ln(x)+c, or Lim(f(x)=negatiive infinite.
    3. However 0<f(x)<1, the contradiction shows: Lim ABS(x*df)=0, in turn, Lim(x*df*f)=0
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  5. #5
    Senior Member Dinkydoe's Avatar
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    I see your problem, if f'(x)= -e^{-ax}....then f(x)=(1/a)\cdot e^{-ax} seems to fit all requirements... (for a>1)

    And \lim_{x\to \infty}(x\cdot f'(x)\cdot f(x))= \lim_{x\to\infty}-(1/a)\cdot x\cdot e^{-2ax}= 0 (using L'hopital)

    But what I think they mean...is that f'_a(x)=-a/x^p is sufficiently close to all functions f'(x) for large x. (that is all f' that fit the requirements...)

    That is, for all \epsilon > 0 we can find N such that |f'(n)-f_a(n)|<\epsilon for n>N.

    So that the same conclusion holds for all f'...
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