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Revision difference : limit of real number sequence
Version 5 Version 4
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.}\, 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
\begin{align}
\pi \;=\, 3.14159265\ldots, \quad 0.636363\ldots, \quad 0.99999\ldots,
\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
\begin{align}
0.9,\;0.99,\;0.999,\;\ldots
\end{align}
which may be also written as
$$1\!-\!\frac{1}{10},\; 1\!-\!\frac{1}{10^2},\; 1\!-\!\frac{1}{10^3},\;\ldots $$
The limit of (7) is 1.\, Actually, if\, $\varepsilon > 0$,\, the distance between 1 and the $n^{\mbox{th}}$ member of (7) is
$$\left|1-\left(1\!-\!\frac{1}{10^n}\right)\right| = \frac{1}{10^n} < \varepsilon,$$
when\, $10^n > \frac{1}{\varepsilon}$,\, i.e. when\, $n > -\lg\varepsilon$.
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. [Not ready...]