What is the limit of the sequence ((e^n) + (e^-n)) / (e^2n - 1)?

[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?

 

[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.

 

[Mark44]

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))

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 e^{2n - 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?

 

[arl146]

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

 

[HallsofIvy]

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

 

[Mark44]

It's not? Is it only true with log then?

It's not true for log in any base.

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

The bottom is (e^2n) - 1

 

[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

 

[Mark44]

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).

 

[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

 

[Mark44]

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

In the future, you should start a new thread for a new problem...

.. 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

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.

 

[arl146]

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

 

[Mark44]

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.

 

[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

 

[Mark44]

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

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

 

[arl146]

it gets closer to 0

 

[Mark44]

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