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

1. ## 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?

2. ## 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?

3. ## Re: limit of a function as it goes to infinity

Originally Posted by alexmahone
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!