|
|
|
Viewing Version
6
of
'limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0'
|
[ view 'limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0'
|
back to history
]
| Title of object: |
limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0 |
| Canonical Name: |
LimitOfDisplaystyleFracsinXxAsXApproaches0 |
| Type: |
Theorem |
| Created on: |
2007-04-24 17:53:53 |
| Modified on: |
2007-05-30 15:37:05 |
| Classification: |
msc:26A03, msc:26A06 |
Revision comment (for changes between this and next version):
| Changes for correction #13002 ('in C'). |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
\newtheorem{thm*}{Theorem}
|
Content:
\begin{thm*}
$$\lim_{x \to 0} \frac{\sin x}{x}=1$$
\end{thm*}
\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.}
\begin{proof}
First, let $\displaystyle 0<x<\frac{\pi}{2}$. Then $0<\cos x<1$. Note also that
\begin{equation}
\label{tan}
x<\tan x.
\end{equation}
Multiplying both \PMlinkescapetext{sides} of this inequality by $\cos x$ yields
\begin{equation}
\label{xcos}
x\cos x<\sin x.
\end{equation}
By \PMlinkname{this theorem}{ComparisonOfSinThetaAndThetaNearTheta0},
\begin{equation}
\label{x}
\sin x<x.
\end{equation}
Combining inequalities (\ref{xcos}) and (\ref{x}) gives
\begin{equation}
\label{sin}
x\cos x<\sin x<x.
\end{equation}
Dividing by $x$ yields
\begin{equation}
\label{squeeze}
\cos x<\frac{\sin x}{x}<1.
\end{equation}
Now let $\displaystyle \frac{-\pi}{2}<x<0$. Then $\displaystyle 0<-x<\frac{\pi}{2}$. Plugging $-x$ into inequality (\ref{squeeze}) gives
\begin{equation}
\label{squeeze-}
\cos (-x)<\frac{\sin (-x)}{-x}<1.
\end{equation}
Since $\cos$ is an even function and $\sin$ is an odd function, we have
\begin{equation}
\label{-squeeze}
\cos x<\frac{-\sin x}{-x}<1.
\end{equation}
Therefore, inequality (\ref{squeeze}) holds for all real $x$ with $\displaystyle 0<|x|<\frac{\pi}{2}$.
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} |
|
|
|
|
|
