How to solve $\lim_{n\to \infty}\sin(1)\times \sin(2)\times\sin(3)\times\ldots\times\sin(n)$ The limits I'm trying to solve are:
$$\lim_{n\to \infty}\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n)$$
$$\lim_{n\to \infty}n\times\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n)$$
For the former limit, my (probably incorrect) solution is that $\sin(1)\times\sin(2)\times\sin(3)\ldots$ are constants, so the limit can be written as 
$$\sin(1)\times\sin(2)\times\sin(3)\times\ldots\times\sin(n-1)\cdot \lim_{n\to \infty}\sin(n)$$
and $\lim_{n\to \infty}\sin(n)$ simply does not exist, because $\sin(n)$ does not settle on a single value when ${n\to \infty}$.
 A: Density (i.e., the irrationality of $\pi$) is not needed. 
Let $f\colon\mathbb R$ be any periodic function with $|f(x)\le 1$ for all $x$ and there exists a closed interval $I$ of length $1$ and a number $q<1$ such that $|f(x)|\le q$ for all $x\in I$.
Then for any $m\in\mathbb N_0$
$$\lim_{n\to \infty}n^mf(1)f(2)\cdots f(n) = 0 $$
To see this let $p$ be a period of $f$ 
Then $I, I+1, \ldots , I+\lceil p\rceil-1$ cover a full period of $f$, hence among any $\lceil p\rceil $ consecutive integers $k+i$, $0\le i<\lceil p\rceil$,  there is at least one with $|f(k+i)\le q$.
As $f(k)|\le 1$ for all other factors, we conclude 
$$\left|\prod_{k=1}^nf(k)\right| \le q^{\lfloor n/\lceil p\rceil\rfloor}<\frac1q\cdot(\sqrt[\lceil p\rceil]q)^n$$
This exponentially small bound implies the claim.
To apply this to the original problem with $f(x)=\sin x$ observe that one may take for example $I=[-\frac12,\frac12]$ and $q=\sin \frac12<\frac12$. 
A: For the first limit, it should be $0$ for the following reason: $\;\{\sin n\mid n\in \mathbf N\}$ is dense in $[-1,1]$. So for any $\varepsilon >0$ there exists $N$ such that $\lvert\, \sin N\,\rvert<\varepsilon$, which implies that for any $n\ge N$,
$$\lvert\,\sin 1\cdot \sin 2\cdots \sin N\cdots\sin n\,\rvert <\varepsilon.$$
A: Hint.
If $$u_n=\sin(1).\sin(2) \dots \sin(n)$$ was having a non vanishing limit $l$ then $$1=\lim\limits_{n \to +\infty} \frac{u_{n+1}}{u_n}= \lim\limits_{n \to +\infty} \sin(n+1)$$ which doesn't make sense as $(\sin(n))_{n \in \mathbb N}$  is dense in $[0,1]$ as you mentionned. So the only potential limit is zero. And indeed the limit is zero again due to the density of $(\sin(n))_{n \in \mathbb N}$ in $[0,1]$.
Regarding the secong sequence $$v_n=n \sin(1).\sin(2) \dots \sin(n)$$ we have with a similar argument that if the limit exists it can only be $0$. Then, again by the density argument, we can find a strictly increasing sequence of integers $(\alpha_n)_{n \in \mathbb N}$ such that $$\vert \sin (\alpha_n) \vert < \frac{1}{2}$$ Hence for $m \ge \alpha_n$ we have $\vert u_m \vert < \frac{1}{2^n}$... Missing some elements to move to the full conclusion...
A: A simple solution is given by considering that:
$$ \left|\sin(a-1)\sin(a)\sin(a+1)\right| \leq \frac{5}{17}  \tag{1}$$
since:
$$ \frac{d}{da}\,\sin(a-1)\sin(a)\sin(a+1) = \frac{1}{2}\cos(a)\left(2+\cos 2-3\cos(2a)\right)\tag{2}$$
so it is not difficult to locate the stationary points of the LHS of $(1)$ and to state:
$$ \left| \sin(1)\cdot \sin(2)\cdot\ldots\cdot\sin(n)\right|\leq \left(\frac{2}{3}\right)^{n-2}\tag{3}$$
from which both limits are trivial.
