$\sin(2\pi nx)$ does not converge for $x \in (0,1/2)$ How to show that $\sin(2 \pi nx)$ does not converge as n goes to infinity?  $x \in (0,1/2)$
 A: Let $x\in (0,1/2)$ and $t\in\mathbb R$. Define $n_j$ as the smallest integer such that $n_j\leqslant (t+j)/x< n_j+1$. Then 
$2\pi n_j x\leqslant 2\pi t+2\pi j \lt 2\pi n_jx+ 2\pi x$, which can be rewritten as 
$$ 2\pi(t-x)<  2\pi n_j x-2\pi j\leqslant  2\pi t  .$$
Choosing $t=(x+1/2)/2$, we get that 
$$ \pi (\frac 12-x)\lt  2\pi n_j x-2\pi j\leqslant  (\frac 12 +x)\pi $$
hence $\sin (2\pi n_j x-2\pi j )=\sin(2\pi n_j x )\geqslant\min\{ \sin(\pi (\frac 12-x)),\sin (\pi (\frac 12+x))\}=:c\gt 0 $. Therefore, the set $I =\{n\mid \sin(2\pi nx )\geqslant c \}$ is infinite. 
Choosing $t= 1/2(x-1/2)$, we get similarly a constant $c'\lt 0$ such that the set $I' =\{n\mid \sin(2\pi nx )\leqslant c' \}$ is infinite. 
A: If $x$ is irrational, there is a stronger result: $\sin(2\pi x \mathbb{Z})$ is dense in $[0,1]$.
Indeed, $\sin(2\pi x \mathbb{Z})= \sin (2\pi x \mathbb{Z} + 2\pi \mathbb{Z})$ and $2\pi x \mathbb{Z}+ 2 \pi \mathbb{Z}$ is a subgroup of $(\mathbb{R},+)$ not of the forme $\alpha \mathbb{Z}$, so it is dense in $\mathbb{R}$. Hence $\sin(2\pi x \mathbb{Z})$ dense in $[0,1]$ since $\sin$ is continuous.
A: $\def\eps{\varepsilon}$Since the statement to be proved is negative (the sequence cannot converge) a fairly elementary proof avoiding any heavy machinery is possible. In spite of the use of $\eps,\delta$ below, the proof involves no limits. In the end a slight computation is needed to arrive at an uncontestable contradiction, but I will try to mention the rather simple line of thought that leads to this argument.
Basically the intuition is that as long as $2\pi x$ stays away from the multiples of $\pi$, the sequnce $n\mapsto\sin(2\pi nx)$ oscillates between $-1$ and $+1$, and can therefore not ultimately stay within a very small neigbourhood of any particular value. To show this rigorously one needs to quantify the fact that $2\pi x$ stays away from multiples of $\pi$; therefore let $\eps=\min(\sin(2\pi x),1-|\cos(2\pi x)|)>0$ so as to ensure that both $\sin(2\pi x)\geq\eps$ and $|\cos(2\pi x)|\leq1-\eps$ (actually one always has $\eps=1-|\cos(2\pi x)|$).
Now imagine that $\lim_{n\to\infty}\sin(2\pi nx)$ exists, then $\sin(2\pi nx)_{n\in\mathbf N}$ would be a Cauchy sequence, which means in particular that for any $\delta>0$ there exists $n$ sufficiently large so that the values $\sin(2\pi(n-1)x),\sin(2\pi nx),\sin(2\pi(n+1)x)$ all lie within $\delta$ of each other. Intuitively oscillating behaviour is not compatible with taking almost the same value thrice in succession (although twice could happen by accident), so I shall deduce a contradiction from this, for which I need $\delta$ to be sufficiently small with respect to $\eps$, in a sense to be made precise. The basic formula used is
$$
  \sin(\alpha\pm y)=\sin\alpha\cos y\pm\cos\alpha\sin y,
$$
with $\alpha=2\pi nx$ and $y=2\pi x$. The final terms have opposite signs for $\sin(2\pi(n-1)x)$ and $\sin(2\pi(n+1)x)$, so if those values are to be within $\delta$ of each other one must have $|\cos\alpha\sin y|<\delta/2$, and given that $\sin y\geq\eps$ this means $|\cos\alpha|<\delta/2\eps$. If we take $\delta$ sufficiently less than $\eps$ this means that $|\cos\alpha|$ must be very small, and $|\sin\alpha|$ therefore very close to $1$. But that means that multiplication by $\cos y<1-\eps$ substantially changes the first term $\sin\alpha\cos y$ with respect to $\sin\alpha=\sin(2\pi nx)$, and because we can have either sign for the second term, this pushes either $\sin(2\pi(n-1)x)$ or $\sin(2\pi(n+1)x)$ even further from $\sin(2\pi nx)$, giving a contradiction. 
Concretely, take $\delta=0.8\eps$, then for $\sin(2\pi(n-1)x)$ and $\sin(2\pi(n+1)x)$ to be less than $\delta$ apart one needs $|\cos\alpha|<0.4$, and therefore $|\sin\alpha|>\sqrt{0.84}>0.9$; since we have $|\cos y|\leq1-\eps$ this means $|\sin\alpha\cos y-\sin\alpha|>0.9\eps>\delta$, which implies that either $|\sin(\alpha+y)-\sin\alpha|>\delta$ or $|\sin(\alpha-y)-\sin\alpha|>\delta$, contradicting the fact that we have a Cauchy sequence. 
