How to show that $\frac{\sin(n)}{n}$ is $1$ as $n \rightarrow 0$? 
Possible Duplicate:
How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? 

How to show that $\frac{\sin(n)}{n}$ 
is $1$ as $n \rightarrow 0$? just hint.
 A: Maclaurin series expansion of $\sin(n)$ is,    
$$\sin(n) = n - \frac{n^3}{3!} +\frac{n^5}{5!}+... $$
Hence,
$$\frac{\sin(n)}{n} = 1-\frac{n^2}{3!} + \frac{n^4}{5!}+...$$
$$\lim_{n\to 0}\frac{\sin(n)}{n} = 1$$
A: First, Prove that $\sin{x}<x<\tan{x}$, when $x\in (0,\frac{\pi}{2})$ By means of drawing a circle, take an arbitary point on the circle with coordinate $A:(\cos{x},\sin{x})$, take $B:(0,1),O:(0,0),C:(\cos{x},0),D:(\sec{x},0)$ 
Obviously We have $\sin{x}=S_{\Delta OAC }$, $x=S_{ OAB}$ where $S_{OAB}$ denotes the area of the circular sector, $\tan{x}=S_{\Delta OAD}$
Also, it's obvious(By drawing this circle) that $S_{\Delta OAC }<S_{ OAB}<S_{\Delta OAD}$, thus\begin{align}\sin{x}<x<\tan{x},\quad(x\in(0,\frac{\pi}{2}))\end{align}
By multiplying $-1$ on each side 
\begin{align}\sin{x}>x>\tan{x},\quad(x\in(-\frac{\pi}{2},0))\end{align}
So we have \begin{align}\cos{x}<\frac{\sin{x}}{x}<1\quad(x\in(-\frac{\pi}{2},\frac{\pi}{2}))\setminus\{0\} \end{align}
Taking the limit will give the result
