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Math Help - Indetermination 1^infinity

  1. #1
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    Indetermination 1^infinity

    [IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
    Hi there. I've found some difficulties on solving this limit:
    \displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}

    I thought of working with the function f(x)=(\displaystyle\frac{x}{x-1})^{x+2}
    And then apply L'Hopital

    This way:
    \displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra  c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}

    And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.
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  2. #2
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    Don't go the L'Hopital route.
    Look at this:
    \left( {\dfrac{x}<br />
{{x + 1}}} \right)^{x + 2}  = \left( {1 - \dfrac{1}<br />
{{x + 1}}} \right)^x \left( {1 - \dfrac{1}<br />
{{x + 1}}} \right)^2
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    Don't go the L'Hopital route.
    Look at this:
    \left( {\dfrac{x}<br />
{{x + 1}}} \right)^{x + 2}  = \left( {1 - \dfrac{1}<br />
{{x + 1}}} \right)^x \left( {1 - \dfrac{1}<br />
{{x + 1}}} \right)^2
    Wouldn't it be more instructive to write it as \displaystyle \left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)?
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Wouldn't it be more instructive to write it as \left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)?
    Not if you know that \displaystyle <br />
\lim _{x \to \infty } \left( {1 + \frac{a}<br />
{{x + b}}} \right)^{cx}  = e^{ac}
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    Not if you know that \displaystyle <br />
\lim _{x \to \infty } \left( {1 + \frac{a}<br />
{{x + b}}} \right)^{cx}  = e^{ac}
    I was thinking that if we let y=x+1 then \displaystyle \left(1-\frac{1}{x+1}\right)^{x+1}=\left(1-\frac{1}{y}\right)^y\to e^{-1}, but both are good.
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  6. #6
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    Quote Originally Posted by Ulysses View Post
    [IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
    Hi there. I've found some difficulties on solving this limit:
    \displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}

    I thought of working with the function f(x)=(\displaystyle\frac{x}{x-1})^{x+2}
    And then apply L'Hopital

    This way:
    \displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra  c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}

    And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.
    If you want to use L'Hospital:

    \displaystyle \lim_{n \to \infty}\left(\frac{n}{n-1}\right)^{n+2} = \lim_{n \to \infty}\left(1 + \frac{1}{n-1}\right)^{n+2}

    \displaystyle = \lim_{n \to \infty}e^{\ln{\left[\left(1+\frac{1}{n-1}\right)^{n+2}\right]}}

    \displaystyle = e^{\lim_{n\to \infty}\ln{\left[\left(1+\frac{1}{n-1}\right)^{n+2}\right]}}

    \displaystyle = e^{\lim_{n \to \infty}(n+2)\ln{\left(1+\frac{1}{n-1}\right)}}

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

    \displaystyle = e^{\lim_{n \to \infty}\frac{\frac{-(n-1)^{-2}}{1 + \frac{1}{n-1}}}{-(n-1)^{-2}}} by L'Hospital's Rule

    \displaystyle = e^{\lim_{n \to \infty}\frac{1}{1+\frac{1}{n-1}}}

    \displaystyle = e^{\lim_{n \to \infty}\frac{n-1}{n}}

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

    \displaystyle =e^1

    \displaystyle = e.


    PHEW!!!
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  7. #7
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    Quote Originally Posted by Ulysses View Post
    [IMG]file:///C:/DOCUME%7E1/PC01/CONFIG%7E1/Temp/moz-screenshot.png[/IMG]
    Hi there. I've found some difficulties on solving this limit:
    \displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}

    I thought of working with the function f(x)=(\displaystyle\frac{x}{x-1})^{x+2}
    And then apply L'Hopital

    This way:
    \displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) \ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\fra  c{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}

    And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.

    It appears that you have dropped the \displaystyle \ln from your last expression. It should be:

    \displaystyle e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle \frac{\ln(\displaystyle \frac{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}

    The limit in this expression is of the form 0/0. Before applying L'H˘pital's rule, let's rewrite the numerator.
    \displaystyle e^{\displaystyle\lim_{x \to\infty}{\displaystyle \frac{\ln(\displaystyle \frac{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}} \displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\ln(x)-\ln(x-1)}\over{\displaystyle{{1}\over{(x+2)}}}}}\right) }

    Apply L'H˘pital's rule:

    \displaystyle\ = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\displaystyle{{1}\  over{x}}-{{1}\over{x-1}}\over{ \displaystyle-{{1}\over{(x+2)^2}}}}}\right) } \displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{{\displaystyle{{x-1-x}\over{x(x-1)}}\over{ \displaystyle-{{1}\over{(x+2)^2}}}}}\right) } \displaystyle = e^{\displaystyle\lim_{x \to\infty}\left(\displaystyle{{(x+2)^2}\over{ x(x-1)}}\right) }=e^1=e

    It works out just fine with L'H˘pital.
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  8. #8
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    Great work. Thank you all guys!
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