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.
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.
Maclaurin series expansion of $\sin(n)$ is,
$\sin(n) = n - \frac{n^3}{3!} +\frac{n^5}{5!}+... $
Hence,
$\frac{\sin(n)}{n} = 1-\frac{n^2}{3!} + \frac{n^4}{5!}+...$
$\lim_{n\to 0}\frac{\sin(n)}{n} = 1$
First, Prove that $\sin{x}
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 } 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