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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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4Show that $\cos{x}<\frac{\sin{x}}{x}<1,x\in(-\frac{\pi}{2},\frac{\pi}{2})$ – Golbez Oct 18 '12 at 15:52
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2As an alternative, you can consider that limit as a derivative. – T. Verron Oct 18 '12 at 15:56
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2T.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. – Elchanan Solomon Oct 18 '12 at 15:57
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1By the way how to formulate arbitrary complex trigonometric polynom? I know that in real form it is $\sum_{n=1}^{k}cos(nx)+isin(nx)$ – Oct 18 '12 at 16:02
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4It 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
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Taylor expansion of $\sin(x)$. – JACKY88 Oct 18 '12 at 16:13
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1Following Golbez' hint, see here. – David Mitra Oct 18 '12 at 16:40
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1"A mathematician’s nightmare is a sequence $n_\varepsilon$ that tends to $0$ as $\varepsilon$ becomes infinite." (P. Halmos) – M. Strochyk Oct 18 '12 at 16:45
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1See also How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?. Another question related to inequality $|\sin x|<|x|$ is this: How to strictly prove $\sin <x$ for $0<x<\frac{\pi}{2}$. – Martin Sleziak Oct 18 '12 at 17:25
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You can actually prove geometrically that $|\sin x -x| < x^2$ for $x$ close to zero, which yields your result. – Thomas Andrews Oct 18 '12 at 18:09
2 Answers
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Maclaurin series expansion of $\sin(n)$ is,
$$\sin(n) = n - \frac{n^3}{3!} +\frac{n^5}{5!}+... $$
Hence,
$$\frac{\sin(n)}{n} = 1-\frac{n^2}{3!} + \frac{n^4}{5!}+...$$
$$\lim_{n\to 0}\frac{\sin(n)}{n} = 1$$
Salech Alhasov
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4This 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
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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
Golbez
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