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Math Help - limit of a function as it goes to infinity

  1. #1
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    limit of a function as it goes to infinity

    I have questions with the following function \displaystyle \begin{align*} \lim_{x \to \infty}({\frac{5x-1}{5x+2})^{5x+1}} \end{align*}. How does one calculate this limit? I know it is e^{-3} but I don't know how to get it. I set the function equal to L and take the natural log of both sides. From there I don't know what to do. I believe you have to use L'hopital rule but I don't see how you could because if you do it like that then you get the answer to be e which is incorrect. What am I doing wrong?

    P.S. Here are my steps

    \displaystyle \begin{align*} \lim_{x \to \infty}({\frac{5x-1}{5x+2})^{5x+1}} \end{align*} = L


    \displaystyle \begin{align*} \lim_{x \to \infty}(5x+1) ln \frac{5x-1}{5x+2} \end{align*} = ln L


    \displaystyle \begin{align*} \lim_{x \to \infty}\frac {ln \frac{5x-1}{5x+2}}{ \frac{1}{5x+1}} \end{align*} = ln L

    Where do I go from here?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: limit of a function as it goes to infinity

    Let 5x=y.

    As x\to\infty, y\to\infty

    \lim_{x\to\infty}\left(\frac{5x-1}{5x+2}\right)^{5x+1}= \lim_{y\to\infty}\left(\frac{y-1}{y+2}\right)^{y+1}

    =\lim_{y\to\infty}\left(\frac{y+2-3}{y+2}\right)^{y+1}

    =\lim_{y\to\infty}\left(1-\frac{3}{y+2}\right)^{y+1}

    Can you proceed?
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  3. #3
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    Re: limit of a function as it goes to infinity

    Quote Originally Posted by alexmahone View Post
    Let 5x=y.

    As x\to\infty, y\to\infty

    \lim_{x\to\infty}\left(\frac{5x-1}{5x+2}\right)^{5x+1}= \lim_{y\to\infty}\left(\frac{y-1}{y+2}\right)^{y+1}

    =\lim_{y\to\infty}\left(\frac{y+2-3}{y+2}\right)^{y+1}

    =\lim_{y\to\infty}\left(1-\frac{3}{y+2}\right)^{y+1}

    Can you proceed?
    Thanks for your help, but I figured it out by multiplying the function inside the natural log by 1/x over 1/x and then doing it. I got the right answer. However, when I did it your way I got e^{\frac {-26}{25}}. How did I go wrong?


    EDIT: I did it again and got e^3, how do I get e^-3?

    EDIT 2: I just found my error. Thank you again for showing me this. This was much easier then the way I did it!
    Last edited by Barthayn; January 15th 2012 at 06:30 PM.
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