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Math Help - Interesting limit problem

  1. #1
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    Interesting limit problem

    We need to show the following,

    Lim X-> 1

    [X(X-1)T+1]^(1/(1-X)) = EXP(-T)

    Obviously we need to apply L'Hopitals rule, but it is not straight forward how to proceed !
    Last edited by frustrated; June 28th 2013 at 07:59 AM.
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  2. #2
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    Re: Interesting limit problem

    I would start by letting y=[x(x-1)T+1]^{\frac{1}{1-x}} and then take the log of both sides and apply the property of logarithms to write it like this:

    ln(y)=\frac{1}{1-x}\ln[x(x-1)T+1]

    And note that you can take the limit of both sides and solve for the limit of y as  x->1

    EDIT, typo:

    \lim_{x->1}ln(y)=\lim_{x->1}\frac{\ln[x(x-1)T+1]}{1-x}
    Last edited by adkinsjr; June 28th 2013 at 08:24 AM.
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  3. #3
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    Re: Interesting limit problem

    Yes, working from that starting point now.
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  4. #4
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    Re: Interesting limit problem

    Quote Originally Posted by frustrated View Post
    We need to show the following,

    Lim X-> 1

    [X(X-1)T+1]^(1/(1-X)) = EXP(-T)
    I simply have no idea what you mean. Have no idea what the function is.

    Please, make the post readable.
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  5. #5
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    Re: Interesting limit problem

    Quote Originally Posted by frustrated View Post
    Yes, working from that starting point now.
    ok, note that I had to edit my post. Now, you know how to deal with the right hand side right? You mentioned L-hopitals rule which sounds right to me since it has the forum \frac{0}{0} at x=1 so try that, and you can solve the equation for \lim_{x->1}y.
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  6. #6
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    Re: Interesting limit problem

    Quote Originally Posted by Plato View Post
    I simply have no idea what you mean. Have no idea what the function is.

    Please, make the post readable.
    He's trying to show that the limit is equal to e^{-T}
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  7. #7
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    Re: Interesting limit problem

    Got it,

    Rewrite as Ln y= Ln[x(x-1)T+1] / (1-x) then apply l'hopitals,

    Rewrite Ln[x(x-1)T+1]=a (1-x)=b

    Lim X->1 =da/dx / db/dx

    da/dx = T(2x-1)/[Tx(x-1)+1]

    db/dx = -1

    da/dx / db/dx = -T(2x-1)/[Tx(x-1)+1] = -T

    ln(y) = -T

    Y= e^-T

    Sorry for many edits !
    Last edited by frustrated; June 28th 2013 at 08:55 AM.
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  8. #8
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    Cool Re: Interesting limit problem

    Quote Originally Posted by frustrated View Post
    Got it,

    Rewrite as Ln y= Ln[x(x-1)T+1] / (1-x) then apply l'hopitals,

    Rewrite Ln[x(x-1)T+1]=a (1-x)=b

    Lim X->1 =da/dx / db/dx

    da/dx = T(2x-1)/[Tx(x-1)+1]

    db/dx = -1

    da/dx / db/dx = -T(2x-1)/[Tx(x-1)+1] = -T

    ln(y) = -T

    Y= e^-T

    Sorry for many edits !


    Yup, i edit my posts constantly too so i do the same thing to everyone
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