This is the correct answer, but I would appreciate it if someone would verify that the work is correct, particularly, is it okay to multiply through by natural log like I did?

$\displaystyle a_n = \frac{e^n+e^{-n}}{e^{2n}-1}$

$\displaystyle \lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{e^n+e^{-n}}{e^{2n}-1} * \frac{\ln}{\ln}$

$\displaystyle L=\lim_{n \to \infty}\frac{n-n}{2n}$

$\displaystyle L=0$