# Sequences - limits

• September 30th 2008, 01:54 PM
symstar
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?

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

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

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

$L=0$
• September 30th 2008, 03:23 PM
kalagota
Quote:

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?

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

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

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

$L=0$

no, it is not correct..
• September 30th 2008, 04:10 PM
symstar
Do you have another suggestion? That was kinda my "last ditch" effort.
• September 30th 2008, 04:50 PM
kalagota
$a_n = \frac{e^n + e^{-n}}{e^{2n}-1} = \frac{e^n + e^{-n}}{e^{n}(e^n-e^{-n})} = \frac{\coth n}{e^n}$
• September 30th 2008, 05:29 PM
symstar
Thanks! Is it possible to do this without using hyperbolic functions though?
• September 30th 2008, 05:51 PM
kalagota
personally, i don't know.. maybe others know it.. hehe
• October 1st 2008, 03:17 AM
Plato
Quote:

Originally Posted by symstar
This is the correct answer,
$a_n = \frac{e^n+e^{-n}}{e^{2n}-1}$

L=0 is correct but your method is not correct.
$\frac{{e^n + e^{ - n} }}
{{e^{2n} - 1}} = \frac{{e^{ - n} + e^{ - 3n} }}
{{1 - e^{ - 2n} }}$

Is the answer now clear?