Convergence of Trig Functions This is not a homework problem, but I'm trying to show that $\sin\frac{1}{n} \rightarrow 0$.
By definition I would start as follows: $|\sin\frac{1}{n} - 0 | = |\sin\frac{1}{n}|$
Since $\sin\frac{1}{n}$ could be negative or positive, I could consider both cases and get an expression $n \geq N(\epsilon)$ involving $\sin^{-1}$. I was wondering if there is a better way of showing that $\sin\frac{1}{n}$ converges to 0?
 A: A common way to prove this is to first show that $|\sin x| \le |x|$.
For fun we  use another approach. Note that if $0 < x < \pi/2$, then $0 < \sin x < 1$ and $0< \cos x< 1$.
Recall the familiar identity
$$\sin 2x =2\sin x\cos x.$$
It is more convenient to rewrite this as 
$$\sin u=2\sin \frac{u}{2}\cos \frac{u}{2}.$$
If $0<u<\pi/2$, we can rewrite this as 
$$\sin \frac{u}{2}=\frac{1}{2}\frac{\sin u}{\cos\frac{u}{2}}.$$
But if $u<\pi/2$, then $\cos\frac{u}{2}>\frac{1}{\sqrt{2}}$, and therefore 
$$\sin \frac{u}{2}<\frac{1}{\sqrt{2}}\sin u.\qquad(\ast)$$
Let $u=1$. Since $\sin 1<1$, we find by using $(\ast)$ that 
$$\sin \frac{1}{2}<\frac{1}{\sqrt{2}}.\qquad (1)$$
Let $u=\frac{1}{2}$. By using $(\ast)$ again, and $(1)$, 
we find that 
$$\sin \frac{1}{4}<\left(\frac{1}{\sqrt{2}}\right)^2.\qquad (2)$$
Let $u=\frac{1}{4}$. By using $(\ast)$ and $(2)$, we find that 
$$\sin\frac{1}{8}<\left(\frac{1}{\sqrt{2}}\right)^3. \qquad (3)$$
Continue. In general we have
$$0<\sin\frac{1}{2^k}<\left(\frac{1}{\sqrt{2}}\right)^k.$$
Thus 
$$\lim_{k\to\infty} \sin\frac{1}{2^k}=0.$$
For $0<x<\pi/2$, the sine function is an increasing function. It follows that
$$\lim_{n\to \infty} \sin\frac{1}{n}=0.$$
