| Version 3 |
Version 2 |
| \begin{thm*} |
\begin{thm*} |
| $$\lim_{x \to 0} \frac{\sin x}{x}=1$$ |
$$\lim_{x \to 0} \frac{\sin x}{x}=1$$ |
| \end{thm*} |
\end{thm*} |
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| \PMlinkescapetext{Note that this entry uses the result} $x<\tan x$ for $\displaystyle 0<x<\frac{\pi}{2}$. \PMlinkescapetext{I have received word that a proof of this fact which uses no calculus will be supplied on PlanetMath. Please let me know when this is completed so that I can edit this entry accordingly.} |
\PMlinkescapetext{Note that this entry uses the result} $x<\tan x$ for $\displaystyle 0<x<\frac{\pi}{2}$. \PMlinkescapetext{I have received word that a proof of this fact which uses no calculus will be supplied on PlanetMath. Please let me know when this is completed so that I can edit this entry accordingly.} |
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| \begin{proof} |
\begin{proof} |
|
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| First, let $\displaystyle 0<x<\frac{\pi}{2}$. Then $0<\cos x<1$. Note also that |
First, let $\displaystyle 0<x<\frac{\pi}{2}$. Then $0<\cos x<1$. Note also that |
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| \begin{equation} |
\begin{equation} |
| \label{tan} |
\label{tan} |
| x<\tan x. |
x<\tan x. |
| \end{equation} |
\end{equation} |
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| Multiplying both \PMlinkescapetext{sides} of this inequality by $\cos x$ yields |
Multiplying both \PMlinkescapetext{sides} of this inequality by $\cos x$ yields |
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| \begin{equation} |
\begin{equation} |
| \label{xcos} |
\label{xcos} |
| x\cos x<\sin x. |
x\cos x<\sin x. |
| \end{equation} |
\end{equation} |
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| By \PMlinkname{this theorem}{ComparisonOfSinThetaAndThetaNearTheta0}, |
By \PMlinkname{this theorem}{ComparisonOfSinThetaAndThetaNearTheta0}, |
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| \begin{equation} |
\begin{equation} |
| \label{x} |
\label{x} |
| \sin x<x. |
\sin x<x. |
| \end{equation} |
\end{equation} |
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| Combining inequalities (\ref{xcos}) and (\ref{x}) yields |
Combining inequalities (\ref{xcos}) and (\ref{x}) yields |
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| \begin{equation} |
\begin{equation} |
| \label{sin} |
\label{sin} |
| x\cos x<\sin x<x. |
x\cos x<\sin x<x. |
| \end{equation} |
\end{equation} |
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| Dividing by $x$ yields |
Dividing by $x$ yields |
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| \begin{equation} |
\begin{equation} |
| \label{squeeze} |
\label{squeeze} |
| \cos x<\frac{\sin x}{x}<1. |
\cos x<\frac{\sin x}{x}<1. |
| \end{equation} |
\end{equation} |
|
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| Now let $\displaystyle \frac{-\pi}{2}<x<0$. Then $\displaystyle 0<-x<\frac{\pi}{2}$. Thus, |
Now let $\displaystyle \frac{-\pi}{2}<x<0$. Then $\displaystyle 0<-x<\frac{\pi}{2}$. Thus, |
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| \begin{equation} |
\begin{equation} |
| \label{squeeze-} |
\label{squeeze-} |
| \cos (-x)<\frac{\sin (-x)}{-x}<1. |
\cos (-x)<\frac{\sin (-x)}{-x}<1. |
| \end{equation} |
\end{equation} |
|
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| Since $\cos$ is an even function and $\sin$ is an odd function, we have |
Since $\cos$ is an even function and $\sin$ is an odd function, we have |
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| \begin{equation} |
\begin{equation} |
| \label{-squeeze} |
\label{-squeeze} |
| \cos x<\frac{-\sin x}{-x}<1. |
\cos x<\frac{-\sin x}{-x}<1. |
| \end{equation} |
\end{equation} |
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| Therefore, inequality (\ref{squeeze}) holds for all real $x$ with $\displaystyle 0<|x|<\frac{\pi}{2}$. |
Therefore, inequality (\ref{squeeze}) holds for all real $x$ with $\displaystyle 0<|x|<\frac{\pi}{2}$. |
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| Since $\cos$ is continuous, $\displaystyle \lim_{x \to 0} \cos x=\cos 0=1$. Thus, by the squeeze theorem, |
Since $\cos$ is continuous, $\displaystyle \lim_{x \to 0} \cos x=\cos 0=1$. Thus, by the squeeze theorem, |
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| \begin{equation} |
\begin{equation} |
| \label{limits} |
\label{limits} |
| 1=\lim_{x \to 0} \cos x \le \lim_{x \to 0} \frac{\sin x}{x} \le \lim_{x \to 0} 1=1. |
1=\lim_{x \to 0} \cos x \le \lim_{x \to 0} \frac{\sin x}{x} \le \lim_{x \to 0} 1=1. |
| \end{equation} |
\end{equation} |
|
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| It follows that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$. |
It follows that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$. |
| \end{proof} |
\end{proof} |
