灵活运用极限加上一些有意思的组合结果我们也可以得到一些相对有意思的结论，比如加上 Dirichlet’s approximation theorem（抽屉原理的简单推论，也可以用连分数展开证明），可以得到如下结果。
Problem 1. (Do not need to complete) Dirichlet’s approximation theorem says that for any irrational number $\alpha$ there are an infinite number of integers $p$ and $q$ so that
$$
\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}
$$
a) Show that when $\alpha$ is irrational, for each rational $r \in \mathbb{Q}$ the set $\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ is finite. What happens when $\alpha$ is rational?
b) Using Dirichlet’s approximation theorem, the fact that $|\sin x| \leq|x|$ and the fact that $\pi$ is irrational, show that there is a subsequence $\left\{\sin n_{k}\right\}_{k=1}^{\infty}$ of $\{\sin n\}_{n=1}^{\infty}$ so that $\lim _{k \rightarrow 0} \sin \left(n_{k}\right)=0 .$
$$
\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}
$$
a) Show that when $\alpha$ is irrational, for each rational $r \in \mathbb{Q}$ the set $\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ is finite. What happens when $\alpha$ is rational?
b) Using Dirichlet’s approximation theorem, the fact that $|\sin x| \leq|x|$ and the fact that $\pi$ is irrational, show that there is a subsequence $\left\{\sin n_{k}\right\}_{k=1}^{\infty}$ of $\{\sin n\}_{n=1}^{\infty}$ so that $\lim _{k \rightarrow 0} \sin \left(n_{k}\right)=0 .$
(Hint: $\mid \sin (m \pi-x)|=| \sin (x) \mid \text { for } m \in \mathbb{N})$
Proof . a)如果$\alpha$是无理数，那么由于$|\alpha-r|>0$，因此这样的$\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ 是有限的。 而如果$\alpha$是有理数，那么当且仅当$|\alpha-r|=0$也就是$\alpha=r$的时候$\left\{(p, q): r=p / q,\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\right\}$ 是无限的
b)直接套用第一题结论加极限的定义
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