| Version 8 |
Version 7 |
| An endless real number sequence |
An endless real number sequence |
| \begin{align} |
\begin{align} |
| a_1,\,a_2,\,a_3,\,\ldots |
a_1,\,a_2,\,a_3,\,\ldots |
| \end{align} |
\end{align} |
| has the real number $L$ as its {\em limit}, if the distance between $L$ and $a_n$ can be made smaller than an arbitrarily small positive number $\varepsilon$ by chosing the \PMlinkescapetext{ordinal number} $n$ of $a_n$ sufficiently great, i.e. greater than a positive number $N$ (the \PMlinkescapetext{size} of which depends on the value of $\varepsilon$); accordingly |
has the real number $L$ as its {\em limit}, if the distance between $L$ and $a_n$ can be made smaller than an arbitrarily small positive number $\varepsilon$ by chosing the \PMlinkescapetext{ordinal number} $n$ of $a_n$ sufficiently great, i.e. greater than a positive number $N$ (the \PMlinkescapetext{size} of which depends on the value of $\varepsilon$); accordingly |
| $$|L-a_n| < \varepsilon \quad \mbox{when} \quad n > N.$$ |
$$|L-a_n| < \varepsilon \quad \mbox{when} \quad n > N.$$ |
| Then we may denote |
Then we may denote |
| \begin{align} |
\begin{align} |
| \lim_{n\to\infty}a_n = L |
\lim_{n\to\infty}a_n = L |
| \end{align} |
\end{align} |
| or equivalently |
or equivalently |
| \begin{align} |
\begin{align} |
| a_n \to L \quad \mbox{as} \quad n \to \infty. |
a_n \to L \quad \mbox{as} \quad n \to \infty. |
| \end{align} |
\end{align} |
|
|
| \textbf{Remark.}\, One should not think, that\, $a_n = L$\, when\, $n = \infty$.\, The symbol ``$\infty$'' \PMlinkescapetext{represents} no number, one cannot set it for the value of $n$.\, It's only a question of allowing $n$ to exceed any necessary |
\textbf{Remark.}\, One should not think, that\, $a_n = L$\, when\, $n = \infty$.\, The symbol ``$\infty$'' \PMlinkescapetext{represents} no number, one cannot set it for the value of $n$.\, It's only a question of allowing $n$ to exceed any necessary |
| value.\\ |
value.\\ |
|
|
| \textbf{Example 1.}\, Using the notation (2) we can write a result |
\textbf{Example 1.}\, Using the notation (2) we can write a result |
| $$\lim_{n\to\infty}\frac{2n}{n\!+\!1} = 2.$$ |
$$\lim_{n\to\infty}\frac{2n}{n\!+\!1} = 2.$$ |
| It's a question of that the real number sequence |
It's a question of that the real number sequence |
| $$\frac{2}{2},\;\frac{4}{3},\;\frac{6}{4},\;\ldots$$ |
$$\frac{2}{2},\;\frac{4}{3},\;\frac{6}{4},\;\ldots$$ |
| has the limit value 2 (e.g. the nine hundred ninety-ninth member\, $\frac{1998}{1000} = 1.998$\, is already ``almost'' 2!).\, For justificating the result, let $\varepsilon$ be an arbitrary positive number, as small as you want.\, Then |
has the limit value 2 (e.g. the nine hundred ninety-ninth member\, $\frac{1998}{1000} = 1.998$\, is already ``almost'' 2!).\, For justificating the result, let $\varepsilon$ be an arbitrary positive number, as small as you want.\, Then |
| \begin{align} |
\begin{align} |
| \left|2-\frac{n}{n\!+\!1}\right| = \left|\frac{2n\!+\!2}{n\!+\!1}-\frac{2n}{n\!+\!1}\right| |
\left|2-\frac{n}{n\!+\!1}\right| = \left|\frac{2n\!+\!2}{n\!+\!1}-\frac{2n}{n\!+\!1}\right| |
| = \left|\frac{2}{n\!+\!1}\right| = \frac{2}{n\!+\!1} < \varepsilon, |
= \left|\frac{2}{n\!+\!1}\right| = \frac{2}{n\!+\!1} < \varepsilon, |
| \end{align} |
\end{align} |
| when $n$ is chosen so big that |
when $n$ is chosen so big that |
| \begin{align} |
\begin{align} |
| n > \frac{2}{\varepsilon}\!-\!1. |
n > \frac{2}{\varepsilon}\!-\!1. |
| \end{align} |
\end{align} |
| The condition (5) is obtained from (4) by solving this inequality for $n$.\, In this case, we have\, |
The condition (5) is obtained from (4) by solving this inequality for $n$.\, In this case, we have\, |
| $N = \frac{2}{\varepsilon}\!-\!1$.\\ |
$N = \frac{2}{\varepsilon}\!-\!1$.\\ |
|
|
| \textbf{Example 2.}\, The so-called decimal expansions, i.e. endless decimal numbers, such as |
\textbf{Example 2.}\, The so-called decimal expansions, i.e. endless decimal numbers, such as |
| \begin{align} |
\begin{align} |
| \pi \;=\, 3.14159265\ldots, \quad 0.636363\ldots, \quad 0.99999\ldots, |
\pi \;=\, 3.14159265\ldots, \quad 0.636363\ldots, \quad 0.99999\ldots, |
| \end{align} |
\end{align} |
| are, as a matter of fact, limits of certain real number sequences.\, E.g. the last of these is related to the sequence |
are, as a matter of fact, limits of certain real number sequences.\, E.g. the last of these is related to the sequence |
| \begin{align} |
\begin{align} |
| 0.9,\;0.99,\;0.999,\;\ldots |
0.9,\;0.99,\;0.999,\;\ldots |
| \end{align} |
\end{align} |
| which may be also written as |
which may be also written as |
| $$1\!-\!\frac{1}{10},\; 1\!-\!\frac{1}{10^2},\; 1\!-\!\frac{1}{10^3},\;\ldots $$ |
$$1\!-\!\frac{1}{10},\; 1\!-\!\frac{1}{10^2},\; 1\!-\!\frac{1}{10^3},\;\ldots $$ |
| The limit of (7) is 1.\, Actually, if\, $\varepsilon > 0$,\, the distance between 1 and the $n^{\mbox{th}}$ member of (7) is |
The limit of (7) is 1.\, Actually, if\, $\varepsilon > 0$,\, the distance between 1 and the $n^{\mbox{th}}$ member of (7) is |
| $$\left|1-\left(1\!-\!\frac{1}{10^n}\right)\right| = \frac{1}{10^n} < \varepsilon,$$ |
$$\left|1-\left(1\!-\!\frac{1}{10^n}\right)\right| = \frac{1}{10^n} < \varepsilon,$$ |
| when\, $10^n > \frac{1}{\varepsilon}$,\, i.e. when\, $n > -\log_{10}\varepsilon$. |
when\, $10^n > \frac{1}{\varepsilon}$,\, i.e. when\, $n > -\log_{10}\varepsilon$. |
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The endless decimal notations (6) and others are, in fact, limit notations --- no finite amount of decimals in them suffices to give their exact values.
|
The endless decimal notations (7) and others are, in fact, limit notations --- no finite amount of decimals in them suffices to give their exact values.
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