IteratedLimitInMathbbR2 2007-08-13 10:45:53 2007-08-19 11:13:07 Definition limit iterated limit % 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} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} Let $f$ be a function from a subset $S$ of\, $\mathbb{R}^2$\, to\, $\mathbb{R}$ and\, $(a,\, b)$\, an accumulation point of $S$. The limits $$\lim_{x\to a}\left(\lim_{y\to b}f(x,\,y)\right) \quad\mbox{and}\quad \lim_{y\to b}\left(\lim_{x\to a}f(x,\,y)\right)$$ are called {\em iterated limits}.\\ \textbf{Example 1.} If\; $\displaystyle f(x,\,y) := \frac{x\sin\frac{1}{x}+y}{x+y}$,\, then \begin{itemize} \item $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = \lim_{x\to0}\sin\frac{1}{x}$ does not exist \item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0}1 = 1$ \item the usual limit $\lim_{(x,y)\to(0,0)}f(x,\,y)$ does not exist. \end{itemize} \textbf{Example 2.} If\; $\displaystyle f(x,\,y) := \frac{x^2}{x^2+y^2}$,\, then \begin{itemize} \item $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = \lim_{x\to0} \frac{x^2}{x^2} = 1$ \item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0} 0 = 0$ \item the usual limit $\lim_{(x,y)\to(0,0)}f(x,\,y)$ again does not exist, \PMlinkescapetext{even} though both of the iterated limits do. \end{itemize} So far we have studied examples that present discontinuity at its point of accumulation. We now expose an illustrative example where such discontinuity can be avoided. \\ \textbf{Example 3.} Consider the function $$f(x,\,y) := \frac{x\sin{x}\cosh{y}+y\cos{x}\sinh{y}}{x^2+y^2};$$ then (we apply \PMlinkname{l'H\^opital's rule}{LHpitalsRule} throughout) \begin{itemize} \item $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = \lim_{x\to 0}\left(\lim_{y\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= \lim_{x\to 0}\frac{x\sin x}{x^2}=\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\cos x=1$ \item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to 0}\left(\lim_{x\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= \lim_{y\to 0}\frac{y\sinh y}{y^2}=\lim_{y\to 0}\frac{\sinh y}{y}=\lim_{y\to 0}\cosh y=1$ \item the usual limit $\lim_{(x,y)\to(0,0)}f(x,\,y)$ exists in this case. An essential reason which assures the continuity of this function, arises from the fact that\, $f(x,\,y) \equiv \Re(\frac{\sin z}{z})$,\, $z = x+iy$,\, i.e. it is the real part of the analytic function \,$w := \frac{\sin z}{z}$\, having the removable singularity at\, $z = 0$ (see the entry complex sine and cosine). \end{itemize}