Limit Theorem: Solving \lim_{n \to \infty} \cos( \frac{2 \pi}{2n - 2} )^n = 1

[andrewm]

I was having a debate with a friend about how to show the following limit.

\lim_{n \to \infty} \cos( \frac{2 \pi}{2n - 2} )^n = 1

I claim that you can just hand-wavingly say that since cosine of 0 is 1, and 1^infinity is 1, the limit is 1. He claims I need to show this using some sort of limit theorem (I don't want to get into delta-epsilons).

Is there a cool limit theorem I can use?I

 

[statdad]

Your latex didn't compile - at least I can't see it. Could/would you repost?

i just tried getting your code - is this your problem?

<br /> \lim_{n \to \infty} \cos( \frac{2 \pi}{2n - 2} )^n = 1<br />

okay, my version of the latex didn't take. is there a general problem with the new server setup?

 

[gel]

I claim that you can just hand-wavingly say that since cosine of 0 is 1, and 1^infinity is 1, the limit is 1.

No, the limit would not be 1 if you replaced the exponent by n^2 (what would it be?). You could take logarithms and apply l'hospital's rule and use cos(0)=1, cos'(0)=0.

 

[andrewm]

No, the limit would not be 1 if you replaced the exponent by n^2 (what would it be?). You could take logarithms and apply l'hospital's rule and use cos(0)=1, cos'(0)=0.

Using l'hospital's rule I can show that \lim_{n \to \infty} \ln f(n) = 0, where f(n) is the original cosine function. If the limit of the \ln is the \ln of the limit, then I am content. Am I misunderstanding what you mean by "take logarithms"?

Thanks for the idea!

 

[gel]

yes. To rigorously finish off the proof you can take the exponential and use the fact that the exponential of a limit equals the limit of the exponentials - because exp is a continuous function.

 

[andrewm]

Excellent, I understand. Thanks.