1. ## limits with e

find the limit of $\displaystyle a_n = \frac{e^n + e^{-n}}{e^{2n} - 1}$ an n approaches infinity

L'Hop's rule doesnt work and dividing the terms by e^2n doesnt work either. I do not know what process should be used.

2. You have only to consider that is ...

$\displaystyle a_{n} = \frac { e^{n} + e^{-n}}{ e^{2n}-1} = e^{-n}\cdot \frac{ e^{n} + e^{-n}}{ e^{n}-e^{-n}}$ (1)

... and then find the limit when n tends to infinity of the two term of (1)... not very difficult...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. $\displaystyle e^{-n}\cdot \frac{ e^{n} + e^{-n}}{ e^{n}-e^{-n}}$
I don't understand how you got that part.

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

... all right?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Another way to do this is to divide both numerator and denominator by $\displaystyle e^{2n}$:
$\displaystyle \frac{e^{-n}+ e^{-3n}}{1- e^{-2n}}$
Now what happens to all negative powers as n goes to infinity?

6. @HallsofIvy
I did that but i guess my algebra was off. Your process works, thanks.