Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$ I am having trouble calculating the following limit:
 $$\lim_{n \to \infty} \sqrt[n]{|\sin n|}\ .$$
 A: If the limit exists, then it should be equal to 1, because $|\sin n|$ is dense in $[0,1]$ and there exists a subsequence $|\sin n_k|$ converging to $1$. Then 
$$\sqrt[n_k]{|\sin n_k|} \to 1.$$
A: Hint:
$\pi$ is not a Liouville number, so there exists $m\in\mathbb{N}$ such that for all $p,q\in\mathbb{Z}$ with $q>1$, we have
$$ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$
This should allow you to keep $\sin n$ away from 0.

Edit: Full Solution:
Let $m$ be as above. So for all $p,q\in\mathbb{Z}$ with $q>1$ we have
$$ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$
Now take $n\in\mathbb{N}$. Take $q_n$ so that $|q_n\pi-n|$ is minimized. Then we have
$$ \frac{\pi}{2} \geq \left| q_n\pi - n\right| \geq \frac{1}{q_n^{m-1}}.$$
Next we note that $|\sin n| = |\sin(q_n\pi-n)|$. Now since $\sin$ is increasing on $[0,\pi/2]$ we have
$$|\sin(q_n\pi-n)|\geq \sin\frac{1}{q_n^{m-1}} \geq \frac{1}{2}\frac{1}{q_n^{m-1}}.$$
Such an estimate holds for each $n$, with $q_n\approx \frac{n}{\pi}$. So now we have
$$\frac{1}{2}\frac{1}{q_n^{m-1}} \leq |\sin n| \leq 1.$$ Now take $n$th roots of everything and let $n\rightarrow\infty$. The LHS goes to 1 so
$$\sqrt[n]{|\sin n|} \rightarrow 1.$$
A: HINT: 
There are arbitrarily large multiples of $\pi$, and there are arbitrarily large odd multiples of $\pi /2$.
Of course, this only happens if we consider real n as opposed to natural n...
