Prove $\cos(n)$ does not converge as $n$ tends to infinity 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?
 A: (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.
A: 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. 
A: 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.    
