| Version current |
Version 6 |
| 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 |
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 |
| $$\lim_{x\to a}\left(\lim_{y\to b}f(x,\,y)\right) \quad\mbox{and}\quad \lim_{y\to b}\left(\lim_{x\to a}f(x,\,y)\right)$$ |
$$\lim_{x\to a}\left(\lim_{y\to b}f(x,\,y)\right) \quad\mbox{and}\quad \lim_{y\to b}\left(\lim_{x\to a}f(x,\,y)\right)$$ |
| are called {\em iterated limits}.\\ |
are called {\em iterated limits}.\\ |
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| \textbf{Example 1.} If\; $\displaystyle f(x,\,y) := \frac{x\sin\frac{1}{x}+y}{x+y}$,\, then |
\textbf{Example 1.} If\; $\displaystyle f(x,\,y) := \frac{x\sin\frac{1}{x}+y}{x+y}$,\, then |
| \begin{itemize} |
\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 $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = \lim_{x\to0}\sin\frac{1}{x}$ does not exist |
| \item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0}1 = 1$ |
\item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0}1 = 1$ |
| \item the usual limit $\lim_{(x,y)\to(0,0)}f(x,\,y)$ does not exist. |
\item the usual limit $\lim_{(x,y)\to(0,0)}f(x,\,y)$ does not exist. |
| \end{itemize} |
\end{itemize} |
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| \textbf{Example 2.} If\; $\displaystyle f(x,\,y) := \frac{x^2}{x^2+y^2}$,\, then |
\textbf{Example 2.} If\; $\displaystyle f(x,\,y) := \frac{x^2}{x^2+y^2}$,\, then |
| \begin{itemize} |
\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 $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = \lim_{x\to0} \frac{x^2}{x^2} = 1$ |
| \item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0} 0 = 0$ |
\item $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = \lim_{y\to0} 0 = 0$ |
| \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. |
\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} |
\end{itemize} |
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| 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. \\ |
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. \\ |
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| \textbf{Example 3.} Consider the function |
\textbf{Example 3.} Consider the function |
| $$f(x,\,y) := \frac{x\sin{x}\cosh{y}+y\cos{x}\sinh{y}}{x^2+y^2};$$ |
$$f(x,\,y) := \frac{x\sin{x}\cosh{y}+y\cos{x}\sinh{y}}{x^2+y^2};$$ |
| then (we apply \PMlinkname{l'H\^opital's rule}{LHpitalsRule} throughout) |
then (we apply \PMlinkname{l'H\^opital's rule}{LHpitalsRule} throughout) |
| \begin{itemize} |
\begin{itemize} |
| \item |
\item |
| $\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = |
$\lim_{x\to 0}\left(\lim_{y\to 0}f(x,\,y)\right) = |
| \lim_{x\to 0}\left(\lim_{y\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= |
\lim_{x\to 0}\left(\lim_{y\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= |
| \lim_{x\to 0}\frac{x\sin x}{x^2}=\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\cos x=1$ |
\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 |
\item |
| $\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = |
$\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y)\right) = |
| \lim_{y\to 0}\left(\lim_{x\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= |
\lim_{y\to 0}\left(\lim_{x\to 0}\frac{x\sin x\cosh y+y\cos x\sinh y}{x^2+y^2}\right)= |
| \lim_{y\to 0}\frac{y\sinh y}{y^2}=\lim_{y\to 0}\frac{\sinh y}{y}=\lim_{y\to 0}\cosh y=1$ |
\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$ |
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\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).
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\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 $w := \frac{\sin z}{z}$ (see the entry complex sine and cosine).
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| \end{itemize} |
\end{itemize} |
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