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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.

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marked as duplicate by Martin Sleziak, AD., rschwieb, Norbert, Jason DeVito Oct 19 '12 at 1:27

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Show that $\cos{x}<\frac{\sin{x}}{x}<1,x\in(-\frac{\pi}{2},\frac{\pi}{2})$ –  Golbez Oct 18 '12 at 15:52
As an alternative, you can consider that limit as a derivative. –  T. Verron Oct 18 '12 at 15:56
T.Verron, that may be a bit circular. You can consider the limit as a derivative, but if you can't prove this limit you can't prove the derivative. –  Isaac Solomon Oct 18 '12 at 15:57
By the way how to formulate arbitrary complex trigonometric polynom? I know that in real form it is $\sum_{n=1}^{k}cos(nx)+isin(nx)$ –  laovultai Oct 18 '12 at 16:02
It is a little confusing to use the notation '$ n \rightarrow 0 $'. We usually reserve $ n $ for natural numbers and $ x $ for real numbers. Hence, for reasons of pedagogy, it is better to write '$ x \rightarrow 0 $'. :) –  Haskell Curry Oct 18 '12 at 16:09

2 Answers 2

up vote 4 down vote accepted

Maclaurin series expansion of $\sin(n)$ is,

$$\sin(n) = n - \frac{n^3}{3!} +\frac{n^5}{5!}+... $$


$$\frac{\sin(n)}{n} = 1-\frac{n^2}{3!} + \frac{n^4}{5!}+...$$

$$\lim_{n\to 0}\frac{\sin(n)}{n} = 1$$

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This reasoning is a bit circular (unless we take the series expansion as the definition of the sine function). The Maclaurin expansion depends on the derivative of the sine function, which depends on the limit we're trying to compute. –  Hans Lundmark Oct 18 '12 at 16:17

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

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