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Math Help - Limits

  1. #1
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    Limits

    lim (x-->infinity) (x/ x+1)^lnx

    Do I change it into an integral?
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\lim_{x\to\infty}\left(\frac{x}{x+1}\  right)^{\ln x}=\lim_{x\to\infty}\left(1-\frac{1}{x+1}\right)^{\ln x}=\lim_{x\to\infty}\left[\left(1-\frac{1}{x+1}\right)^{-(x+1)}\right]^{-\frac{ln x}{x+1}}=
    \displaystyle=e^{\displaystyle-\lim_{x\to\infty}\frac{\ln x}{x+1}}=e^{\displaystyle-\lim_{x\to\infty}\frac{1}{x}}=e^0=1
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  3. #3
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    I'm a bit confused... so

    I got lost after the 3rd step...and also what does e equal to?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    I'm a bit confused... so

    I got lost after the 3rd step...and also what does e equal to?
    e = \lim_{n \to \infty} \left(1 + \frac 1n \right)^n \approx 2.71828183

    but that's not what red_dog used here. he used the fact that \frac 1e = \lim_{n \to \infty} \left( 1 - \frac 1n \right)^n

    thus \lim_{n \to \infty} \left( 1 - \frac 1n \right)^{-n} = \lim_{n \to \infty} \frac 1 { \left( 1 - \frac 1n \right)^{n}} = \frac 1{ \lim_{n \to \infty} \left( 1 - \frac 1n \right)^{n}} = e

    here you could replace n with x + 1


    i hope you understand the rest, if not say so. (he used L'Hopital's rule to evaluate the limit in the power)
    Last edited by Jhevon; October 7th 2007 at 10:04 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by circuscircus View Post
    I'm a bit confused... so

    I got lost after the 3rd step...and also what does e equal to?
    How could you be expected to do this limit if you don't know the definition of e?? (Seeing as ln(x) is the inverse function to e^x.)

    -Dan
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