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Math Help - How do I find this limit

  1. #1
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    How do I find this limit

    Limit as x approaches infinity [(x+1)/(x-1)]^x

    I divided everything by an x. I see that the numerator would be e; but not too sure what to do or if that is the right process.

    Help
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  2. #2
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    Quote Originally Posted by Nichelle14
    Limit as x approaches infinity [(x+1)/(x-1)]^x

    I divided everything by an x. I see that the numerator would be e; but not too sure what to do or if that is the right process.

    Help
    I am going to find the limit of,
    \lim_{x\to\infty} \left( 1-\frac{1}{x} \right). I will assume that it exists, (I will leave that to you to prove).

    Begin by saying the limit exists,
    \lim_{x\to\infty}\left(1- \frac{1}{x} \right)=L
    Thus, (using the concept of countinous functions)
    \ln L=\lim_{x\to\infty} \ln \left(1-\frac{1}{x} \right)^x
    Using, the properties of logarithms,
    \ln L=\lim_{x\to\infty} x\ln (1-1/x)
    Since, |1-1/x|<1 it is expressable as a Taylor polynomial,
    Thus,
    \ln L=\lim_{x\to\infty} x\left( -(1/x)-\frac{(1/x)^2}{2}-\frac{(1/x)^3}{3}-... \right)
    Open parantheses,
    \ln L=\lim_{x\to\infty} -1-\frac{1}{2x}-\frac{1}{3x^2}-...
    Since, each term with and "x" gives a zero you have,
    \ln L=-1-0-0-0...
    Thus,
    \ln L=-1
    Finally,
    L=\frac{1}{e}
    ---
    Thus, given,
    <br />
\lim_{x\to\infty} \left( \frac{x+1}{x-1} \right)^x
    Divide by "x" as you said,
    \lim_{x\to\infty}  \frac{(1+1/x)^x}{(1-1/x)^x}
    Since the numerator and denominator (non-zero) both exist you can conclude that the limit is,
    \frac{e}{\frac{1}{e}}=e^2
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  3. #3
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    Hello, Nichelle14!

    You were on the right track . . .


    \lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^x

    Inside the parentheses, divide top and bottom by x:\;\;\lim_{x\to\infty} \frac{\left(1 + \frac{1}{x}\right)^x} {\left(1 - \frac{1}{x} \right)^x}

    . . So we have: . \displaystyle{\frac{\lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^x} {\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x} }


    You recognized that the numerator is e . . . good!

    But did you know that the denominator is e^{-1} ? **


    Therefore, the limit is: . \frac{e}{e^{-1}} \:=\:\boxed{e^2}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    A very handy theorem: . \lim_{x\to\infty}\left(1 + \frac{k}{x}\right)^x\;=\;e^k

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