differentiable functions are continuousDifferentiableFunctionsAreContinuous2004-09-09 14:12:312004-10-21 03:54:55Theoremdifferentiable function% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\sR}[0]{\mathbb{R}}
\newcommand{\sC}[0]{\mathbb{C}}
\newcommand{\sN}[0]{\mathbb{N}}
\newcommand{\sZ}[0]{\mathbb{Z}}
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand*{\norm}[1]{\lVert #1 \rVert}
\newcommand*{\abs}[1]{| #1 |}
\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}\begin{prop}
Suppose $I$ is an open interval on $\mathbb{R}$,
and $f\colon I\to \sC$ is differentiable at $x\in I$. Then
$f$ is continuous at $x$. Further, if $f$ is differentiable on $I$,
then $f$ is continuous on $I$.
\end{prop}
\begin{proof}
Suppose $x\in I$. Let us show that
$f(y)\to f(x)$, when $y\to x$. First, if $y\in I$ is distinct to $x$,
then
$$
f(x)-f(y) = \frac{f(x)-f(y)}{x-y} (x-y).
$$
Thus, if $f'(x)$ is the derivative of $f$ at $x$, we have
\begin{eqnarray*}
\lim_{y\to x} f(x)-f(y) &=& \lim_{y\to x} \frac{f(x)-f(y)}{x-y} (x-y) \\
&=& \lim_{y\to x} \frac{f(x)-f(y)}{x-y}\ \lim_{y\to x} (x-y) \\
&=& f'(x)\ 0 \\
&=& 0,
\end{eqnarray*}
where the second equality is justified since both limits on the second line
exist. The second claim follows since $f$ is continuous on $I$ if and only
if $f$ is continuous at $x$ for all $x\in I$.
\end{proof}