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Math Help - Applying l'Hospital's rule to find a limit

  1. #1
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    Applying l'Hospital's rule to find a limit

    I couldn't find the following limit:


    limxx((1+1x)xe).
    Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
    limx(1+1x)xe1/x.
    Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

    Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks
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  2. #2
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    Re: Applying l'Hospital's rule to find a limit

    Quote Originally Posted by skybluesea2010 View Post
    I couldn't find the following limit:


    limxx((1+1x)xe).
    Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
    limx(1+1x)xe1/x.
    Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

    Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks
    Sorry but this is unreadable.
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    Re: Applying l'Hospital's rule to find a limit

    Quote Originally Posted by skybluesea2010 View Post
    I couldn't find the following limit:


    limxx((1+1x)xe).
    Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
    limx(1+1x)xe1/x.
    Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

    Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks
    Presumably this means:

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

    CB
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    Re: Applying l'Hospital's rule to find a limit

    Exactly, this that's the expression i want to find its limit when x tends to infinity
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    Re: Applying l'Hospital's rule to find a limit

    Quote Originally Posted by CaptainBlack View Post
    Presumably this means:

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

    CB
    Quote Originally Posted by skybluesea2010 View Post
    Exactly, this that's the expression i want to find its limit when x tends to infinity
    Then you apply L'Hopital's rule to:

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