# 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?

-
Do you know that $|\sin x| \leq |x|$ for all $x$? – Antonio Vargas Mar 5 '12 at 6:37
Have you thought about using the sequential definition of continuity,i.e., $x_n\rightarrow x$ then...... . If you use/assume continuity of sinx, this would help. – AQP Mar 5 '12 at 6:40
@AntonioVargas: Could you give me a hint as to how to show that fact? – user26139 Mar 5 '12 at 7:09
I think I see it. Since $0 \leq |\sin x| \leq 1$ and $|x| \in (0, \infty)$, $\frac{|\sin x|}{|x|} \leq 1$ if $x \neq 0$. By rearrangement, we get that $|\sin x| \leq |x|$. – user26139 Mar 5 '12 at 7:22
@user26139: I assume that $n$ ranges over the positive integers. Then you don't need to worry about $\sin\frac{1}{n}$ being negative, since $0<\frac{1}{n}\le 1$, and the sine of anything between $0$ and $\pi/2$ is positive. – André Nicolas Mar 5 '12 at 8:50

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 $$\lim_{n\to \infty} \sin\frac{1}{n}=0.$$