# Homework Help: Finding limit of a sequence

1. May 1, 2012

### arl146

The sequence is:

((e^n) + (e^-n)) / (e^2n - 1)

I don't know how to find this limit. Am I supposed to take the natural log of each term? If so you end up with:

(n*ln(e) + (-n)*ln(e)) / (2n*ln(e) - ln(1))

Which all the ln(e) are just equaling 1 so it becomes:

(n-n) / (2n - ln(1))

And since the top is 0, the whole limit equals 0. Correct?

2. May 1, 2012

### thebrainstorm

I would advice you to refer the elementary calculus book by I.A.Maron where finding the limit of a sequence has been clearly explained.

3. May 1, 2012

### Staff: Mentor

This is incorrect. It is NOT true that ln(A + B) = ln(A) + ln(B).

Also, what is the exponent on e in the denominator? Is the expression e2n or e2n - 1?

4. May 1, 2012

### arl146

It's not? Is it only true with log then? The bottom is (e^2n) - 1

5. May 1, 2012

### HallsofIvy

This has nothing to do with logarithms. Divide both numerator and denomimator by $e^{2n}$.

6. May 1, 2012

### Staff: Mentor

It would be good for you to brush up on the properties of logs, and probably exponents as well.

7. May 3, 2012

### arl146

Oh ok i got it. I have no idea what I was thinking with this. Sorry! When you divide by e^(2n) you end up with:

((1/e^n) + (1/e^n)) / (1-(1/e^2n))

And the top is 0 when n approaches infinity so that's how they got 0

8. May 3, 2012

### Staff: Mentor

You got the right answer, but there is one error in your problem. After dividing numerator and denominator by e^(2n) -- in effect, you are multiplying the whole fraction by 1 -- you get
$$\frac{e^{-n} + e^{-3n}}{1 - e^{-2n}}$$

Your error was in the 2nd term in the numerator. You had 1/e^n (same as e^(-n)) instead of 1/e^(3n) (same as e^(-3n).

9. May 3, 2012

### arl146

Yes I just caught that now. Thanks for catching that!

Not sure if I should create a new post but i have a similar question .. If the sequence is 2^n / (3^(n+1)) how do I begin this one .. Usually I'd use l'hopitals or divide by the highest exponent in the denominator, like we just did. But idk how this one works

10. May 3, 2012

### Staff: Mentor

In the future, you should start a new thread for a new problem...
You can write it as
$$(1/3)\frac{2^n}{3^n}$$
Do you see where to go from here? You don't need L'Hopital's Rule for this.

11. May 3, 2012

### arl146

Would I take the natural log of the (2/3)^n ?

12. May 3, 2012

### Staff: Mentor

No, not at all. As n gets large, what happens to (2/3)^n? This is a very simple limit.

If the answer is not obvious to you, look at the first few terms in the sequence.

13. May 3, 2012

### arl146

as n gets large, (2/3)^n gets large too .. ohhhhh nevermind i was looking at it completely wrong again. duh. sorry for wasting your time on these 2 simple problems! i saw that it got closer to 0 but i have no idea what i was thinking before

14. May 3, 2012

### Staff: Mentor

Are you saying that (2/3)^n gets large or gets close to zero?

15. May 3, 2012

### arl146

it gets closer to 0

16. May 4, 2012

### Staff: Mentor

Yes. If |a| < 1, then $\lim_{n \to \infty} a^n = 0$