The set $\{\frac{\varphi(n)}n:n\in \Bbb N\}$ Let $f(n)=\varphi(n)/n$, where $\varphi$ is the totient function. Since $0<\varphi(n)\le n$ and
$$\lim_{n\to \infty}f(p_n)=1$$
(where $\{p_n\}$ is the increasing sequence of primes) and
$$\lim_{n\to\infty}f(n\#)=0$$
we know that $\limsup f(n)=1$ and $\liminf f(n)=0$. But is the set $\{f(n):n\in\Bbb N\}$ dense in $[0,1]$?
Warning: this is not a problem from a book, so it might be very hard (honestly, I have no idea).

Yes. As $\sum_p\frac{1}{p}=+\infty$, we get $$\prod_p f(p)=0.$$ Now choose $a\in[0,1]$ and a prime $p_n$ such that $\frac{p_n-1}{p_n}\geq1-a$. Consider the sequence $$u_n=\prod_{k=0}^nf(p_{n+k})=\frac{\varphi\left(\prod_{k=0}^np_{n+k}\right)}{\prod_{k=0}^np_{n+k}}=f\left(\prod_{k=0}^np_{n+k}\right).$$ Then $u_n\to0$ and the difference between consecutive terms is at most $a$, so we're done. 
