1. ## Sequences - limits

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?

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

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

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

$\displaystyle L=0$

2. Originally Posted by symstar
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?

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

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

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

$\displaystyle L=0$
no, it is not correct..

3. Do you have another suggestion? That was kinda my "last ditch" effort.

4. $\displaystyle a_n = \frac{e^n + e^{-n}}{e^{2n}-1} = \frac{e^n + e^{-n}}{e^{n}(e^n-e^{-n})} = \frac{\coth n}{e^n}$

5. Thanks! Is it possible to do this without using hyperbolic functions though?

6. personally, i don't know.. maybe others know it.. hehe

7. Originally Posted by symstar
$\displaystyle a_n = \frac{e^n+e^{-n}}{e^{2n}-1}$
$\displaystyle \frac{{e^n + e^{ - n} }} {{e^{2n} - 1}} = \frac{{e^{ - n} + e^{ - 3n} }} {{1 - e^{ - 2n} }}$