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Math Help - proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

  1. #1
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    proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

    Hi everyone,

    Stuck on the following proof, any help would be appreciated! (Calculus a Complete Course, Adams, 3rd Ed., Ch3 Ch Review, in case anyone has the text)

    I *think* I have part a) ok

    a) Show that the fuction f(x) = x^x is strictly increasing on [e^-1, if[

    y = x^x
    ln(y) = x*ln(x)
    d/dx => 1/y * y' = x * 1/x + ln(x)
    y' = y(1+ln(x)) = x^x *(1+ln(x))
    if x = 1/e, ln(x) = -1,
    therefore y' = y (1 -1)= 0
    for n > e^-1
    y' = n^n * (1+ln(n)) > 0
    induction, k = n0, for k+1
    y' = (k+1)^(k+1)*(1+ln(k+1)) > 0
    Derivative is always greater than zero so strictly increasing

    I am lost on b)

    b) If g is the inverse function to f of part (a), show that

    lim y-> inf g(y)ln(ln(y))/ln(y) = 1

    hint: start with the eq. y = x^x and take the ln of both sides twice

    y = x^x
    ln(ln(y)) = ln(ln(x^x))

    Now what? I spent ages trying to figure this out and got no where

    Cheers?
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  2. #2
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    Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

    From y=x^x, we get ln(ln(y))=ln(x\cdot ln(x))=ln(x)+ln(ln(x)). Now replace the limit as y goes to infinity, by the limit as x goes to infinity with the function written in terms of x.
    Thanks from tamanous
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  3. #3
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    Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

    Gah! I was missing such a simple step! I totally forgot about the rule for the summation of logarithms (fail...)

    I think I got it now.

    y = x^x
    => ln(ln(y)) = ln(x*ln(x)) = ln(x) + ln(ln(x)), using ln(xy) = ln(x) + ln(y)
    as y -> inf x -> inf since y=x^x

    Therefore,
    use lim x->inf, subbing in x^x
    lim lim x->inf, g(y)ln(ln(y))/ln(y) = g(x^x)[ln(x)+ln(ln(x))]/(x*ln(x))
    since g is the inverse of f, g(x^x) = x
    Therefore,
    g(x^x)[ln(x)+ln(ln(x))]/(x*ln(x)) = x*[ln(x)+ln(ln(x))]/(x*ln(x)) = [ln(x)+ln(ln(x))]/ln(x)
    = ln(x)/ln(x) + ln(ln(x))/ln(x)
    = 1 + ln(ln(x))/ln(x)

    as x -> inf, ln(ln(x))/(ln(x)) = 0, since ln(x) increases faster than ln(ln(x)) as x->inf, [d/dx ln(x) = 1/x, d/dx ln(ln(x)) = 1/(x*ln(x))]

    therefore as x = inf, 1 + ln(ln(x))/ln(x) = 1

    Thanks a lot! Much love to these forums!!
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  4. #4
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    Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

    what about the g(x) the inverse of y = x^x ?
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  5. #5
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    Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

    Quote Originally Posted by MINOANMAN View Post
    what about the g(x) the inverse of y = x^x ?
    Property of inverse functions: inverse_f(f(x)) = x for all x in the domain of f.
    Therefore, since g is the inverse of f, g(f(x)) = g(x^x) = x
    Inverse function - Wikipedia, the free encyclopedia <- "Inverses and composition"

    Seems ok to me. Did I do something wrong?
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