<?xml version="1.0" encoding="UTF-8"?>

<record version="7" id="9255">
 <title>limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0</title>
 <name>LimitOfDisplaystyleFracsinXxAsXApproaches0</name>
 <created>2007-04-24 17:53:53</created>
 <modified>2007-08-18 04:31:17</modified>
 <type>Theorem</type>
 <creator id="1863" name="Wkbj79"/>
 <author id="1863" name="Wkbj79"/>
 <classification>
	<category scheme="msc" code="26A03"/>
	<category scheme="msc" code="26A06"/>
 </classification>
 <related>
	<object name="ComparisonOfSinThetaAndThetaNearTheta0"/>
	<object name="SincFunction"/>
	<object name="DerivativesOfSinXAndCosX"/>
	<object name="DerivativesOfSineAndCosine"/>
 </related>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}

\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}

\newtheorem{thm*}{Theorem}
</preamble>
 <content>\begin{thm*}
$$\lim_{x \to 0} \frac{\sin x}{x}=1$$
\end{thm*}

\PMlinkescapetext{Note that this entry uses the result} $x&lt;\tan x$ for $\displaystyle 0&lt;x&lt;\frac{\pi}{2}$.  \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.}

\begin{proof}

First, let $\displaystyle 0&lt;x&lt;\frac{\pi}{2}$.  Then $0&lt;\cos x&lt;1$.  Note also that

\begin{equation}
\label{tan}
x&lt;\tan x.
\end{equation}

Multiplying both \PMlinkescapetext{sides} of this inequality by $\cos x$ yields

\begin{equation}
\label{xcos}
x\cos x&lt;\sin x.
\end{equation}

By \PMlinkname{this theorem}{ComparisonOfSinThetaAndThetaNearTheta0},

\begin{equation}
\label{x}
\sin x&lt;x.
\end{equation}

Combining inequalities (\ref{xcos}) and (\ref{x}) gives

\begin{equation}
\label{sin}
x\cos x&lt;\sin x&lt;x.
\end{equation}

Dividing by $x$ yields

\begin{equation}
\label{squeeze}
\cos x&lt;\frac{\sin x}{x}&lt;1.
\end{equation}

Now let $\displaystyle \frac{-\pi}{2}&lt;x&lt;0$.  Then $\displaystyle 0&lt;-x&lt;\frac{\pi}{2}$.  Plugging $-x$ into inequality (\ref{squeeze}) gives

\begin{equation}
\label{squeeze-}
\cos (-x)&lt;\frac{\sin (-x)}{-x}&lt;1.
\end{equation}

Since $\cos$ is an even function and $\sin$ is an odd function, we have

\begin{equation}
\label{-squeeze}
\cos x&lt;\frac{-\sin x}{-x}&lt;1.
\end{equation}

Therefore, inequality (\ref{squeeze}) holds for all real $x$ with $\displaystyle 0&lt;|x|&lt;\frac{\pi}{2}$.

Since $\cos$ is continuous, $\displaystyle \lim_{x \to 0} \cos x=\cos 0=1$.  Thus,

\begin{equation}
\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.
\end{equation}

By the squeeze theorem, it follows that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$.
\end{proof}

Note that the above limit is also valid if $x$ is considered as a complex variable.</content>
</record>
