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?