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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.

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    Show that $\cos{x}<\frac{\sin{x}}{x}<1,x\in(-\frac{\pi}{2},\frac{\pi}{2})$2012-10-18
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    As an alternative, you can consider that limit as a derivative.2012-10-18
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    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.2012-10-18
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    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)$2012-10-18
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    It is a little confusing to use the notation '$ 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 $'. :)2012-10-18
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    Taylor expansion of $\sin(x)$.2012-10-18
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    Following Golbez' hint, see [here](http://math.stackexchange.com/questions/111645/how-do-i-get-cos-theta-lt-frac-sin-theta-theta-lt-1/111659#111659).2012-10-18
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    "A mathematician’s nightmare is a sequence $n_\varepsilon$ that tends to $0$ as $\varepsilon$ becomes infinite." (P. Halmos)2012-10-18
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    See also [How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?](http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1). Another question related to inequality $|\sin x|<|x|$ is this: [How to strictly prove $\sin for $0](http://math.stackexchange.com/questions/125298/how-to-strictly-prove-sin-xx-for-0x-frac-pi2).2012-10-18
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    You can actually prove geometrically that $|\sin x -x| < x^2$ for $x$ close to zero, which yields your result.2012-10-18

2 Answers 2

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

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

Hence,

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

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

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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.2012-10-18
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First, Prove that $\sin{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 }, thus\begin{align}\sin{x}

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