# Superior limit of $cos(nx)$

Let $x$ be a real number, and $({x}/{\pi})\notin \mathbb{Q}.$ Consider sequence $a_n=cos(nx),(n\in \mathbb{N}),$ then we can see that $\displaystyle\lim_{n\to\infty} a_n$ does not exit. But since $a_n$ is bounded, it has a superior limit and a inferior limit. I guess that the superior limit of $a_n$ is a positive number and the inferior limit is $0$, but cannot prove it strictly.

Question: Prove that $$\displaystyle\overline{\lim_{n \to \infty}}cos(nx)>0,$$ and$$\underline{\lim_{n \to \infty}}cos(nx)=0.$$ edit: @Lord Shark the Unknown points out in his answer that $\displaystyle\underline{\lim_{n\to \infty}} cos(nx)$ is actually a negative number(accurately speaking, $\displaystyle\underline{\lim_{n\to \infty} }cos(nx)=-1$.)

• The latter isn't true when $x=2\pi$. – Lord Shark the Unknown Oct 22 '17 at 6:45
• @Lord Shark the Unknown: Right, I'll edit my question. :) – painday Oct 22 '17 at 6:48

Let's think of this geometrically. $\cos nx$ is the $x$-coordinate of the point $P_n=(\cos nx,\sin nx)$. The points $P_n$ lie on the unit circle. Each is obtained by rotating the previous point $x$ radians anticlockwise.
We get a dichotomy. When $x$ is a rational multiple of $2\pi$, the sequence $(P_n)$ is periodic. It takes a finite number of values, each infinitely often. These points form the vertices of a regular $m$-gon, including the point $(1,0)$. Therefore $\limsup\cos nx=1$. If $m$ is even, then another vertex is $(-1,0)$ leading to $\liminf\cos nx=-1$. If $m$ is odd, then $\liminf\cos nx>-1$.
The other case is when $x$ is not a rational multiple of $2\pi$. In this case the sequence $(P_n)$ is dense on the unit circle. Then $\limsup\cos nx=1$ and $\liminf\cos nx=-1$.