Let $f(x)$ be continuous on $R$. Suppose that $f$ is periodic with the minimal postive period $\mu>0$, $\mu$ is irrational). Show that $\lim_{n\to\infty} f(n)$ does not exist.
I do know that $\{n+m\mu;\ n,m\in Z\}$ is dense, but for $x=\lim_{k\to\infty}(n_k+m_k\mu)$, I could not ensure that $n$ is positive, and so could not get a contradiction with the non-existence of $\lim_{n\to\infty} f(n)$.
A related fact is that $\{0+m\mu \mod 1\}_{m\in \mathbb{N}}$ is dense in $[0,1)$. This means (check this) there is a subsequence $m_k \uparrow \infty$ such that
$$ {m_k \mu - [m_k\mu] \downarrow 0} $$ where $[x]$ denotes the integer part. Call $n_k:= [m_k\mu]$ and note $n_k\rightarrow \infty$.
And $f(n_k) = f(m_k\mu-\epsilon_k) = f(-\epsilon_k) \rightarrow f(0)$ by continuity, where $\epsilon_k$ is a quantity going to 0.
In fact, we found subsequence conveging to $f(0)$ but you can find a subsequence converging to $f(x_0)$ for any $x_0$. This is because every orbit of $+\mu$ on $[0,1)$ is dense, i.e., for any $x_0$, the set $\{x_0+m\mu \mod 1\}_{m\in\mathbb{N}}$ is dense in $[0,1)$. You can find a subsequence $m_j\uparrow \infty$ such that $$x_0+m_j\mu - [x_0+m_j\mu] \downarrow 0$$ and repeating as above, go on to show that $f(n_j) \rightarrow f(x_0)$.
