3

How do I go about proving that $\lim\limits_{n \to \infty} \cos(n)$ does not exist where $n\in \mathbb{N}$ using an $\epsilon-N$ style method?

elDin0
  • 1,396
  • 1
    Hint: each interval $(n\pi-1/2,n\pi+1/2)$ contains an integer. – David Mitra Feb 08 '15 at 12:41
  • 4
    The hint is for the question in your title. You would not go about proving the claim in the post body. – David Mitra Feb 08 '15 at 12:51
  • @DavidMitra You scared me. I read only the question in the body, then saw your comment as if the statement was true and I was thinking I had been missing on some really fundamental for many years. – Git Gud Feb 08 '15 at 12:52
  • 5
    I don't get it. Question body and title ask two different things! – rubik Feb 08 '15 at 12:59
  • Some related posts: http://math.stackexchange.com/questions/697145/limit-of-sequence-s-n-cosn, http://math.stackexchange.com/questions/843664/limit-points-of-cos-n, http://math.stackexchange.com/questions/136897/what-is-limsup-limits-n-to-infty-cos-n-when-n-is-a-natural-number, http://math.stackexchange.com/questions/27218/prove-that-the-limit-of-sin-n-as-n-rightarrow-infty-does-not-exist – Martin Sleziak Feb 08 '15 at 14:00
  • Also related. Martin Sleziak posted a better one though. – Caleb Stanford Jul 17 '16 at 02:01

3 Answers3

11

Suppose $\{\cos(n)\}$ converges to $\alpha$. Then $\{\sin(n)\}$ also converges to some $\beta$ such that $$\tag{1}\alpha^2+\beta^2=1$$ since $\sin^2(n)+\cos^2(n)=1$. On the other hand, (see here)$$\cos(n+1)=\cos(n)\cos(1)-\sin(n)\sin(1)$$ and $$\sin(n+1)=\sin(n)\cos(1)+\cos (n)\sin(1)$$ which implies that $$\tag{2}\alpha=\alpha\cos(1)-\beta\sin(1)$$ and $$\tag{3}\beta=\beta\cos(1)+\alpha\sin(1).$$ But we can see that $(1)$, $(2)$ and $(3)$ contradict to each other.

Paul
  • 19,140
10

(Similar to Paul's answer)

Suppose towards contradiction that $\cos(n) \to r$ for some $r$. Then $\cos(n+1) \to r$ also. Moreover, the identity $$ \sin(n) = \frac{\cos(n)\cos(1) - \cos(n + 1)}{\sin(1)} $$ implies that $\sin(n)$ converges, so say it converges to some real number $s$.

Taking the limit of the identity $\cos^2 n + \sin^2 n = 1$, we get that $r^2 + s^2 = 1$. On the other hand, taking the limit of the identity $\cos((n+1) - n) = \cos (n+1) \cos (n) + \sin(n+1) \sin (n)$, we get $r^2 + s^2 = \cos 1$. Hence, $\cos 1 = 1$, which is a contradiction.

Caleb Stanford
  • 45,895
  • And why do you post an answer that is similar to an answer posted one and a half year ago? – miracle173 Jul 17 '16 at 02:11
  • 3
    @miracle173 I wrote it elsewhere, so it was no effort to transfer it to here. In my judgment, it added to the existing answers. No need to be so confrontational. There's a lot of actually bad content you could be complaining about. – Caleb Stanford Jul 17 '16 at 02:36
  • +1. This answer is certainly better and is lot easier to understand – Akash Karnatak Sep 15 '19 at 11:42
0

Use periodicity of $\cos x$: assume $\exists$ L s.t. $|f(x) - L| <0.1 \ \forall x>x_0$. Let's take $x_0= \frac{\pi}{2} + 2 \pi k$ s.t. $\cos x_0 = 0$, so we get $-0.1<L<.1$.

Then, certainly,$-0.1<\cos (x_0 +\frac{\pi}{2})-L|<0.1 \to -0.1<-1-L<0.1$ and we find that $-1.1<L<-0.9$, which is a contradiction. Hence, L doesn't exist.

Alex
  • 19,262