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Thread: 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 $\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}({\frac{5x-1}{5x+2})^{5x+1}} \end{align*}$. How does one calculate this limit? I know it is $\displaystyle e^{-3}$ 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 $\displaystyle e$ which is incorrect. What am I doing wrong?

    P.S. Here are my steps

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


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


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

    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 $\displaystyle 5x=y$.

    As$\displaystyle x\to\infty$, $\displaystyle y\to\infty$

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

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

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

    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 $\displaystyle 5x=y$.

    As$\displaystyle x\to\infty$, $\displaystyle y\to\infty$

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

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

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

    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 $\displaystyle e^{\frac {-26}{25}}$. 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; Jan 15th 2012 at 06:30 PM.
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