| Version 2 |
Version 1 |
| 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 limit, if the distance of $a_n$ from $L$ 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 $N$ (which depends on the value of $\varepsilon$); accordingly |
has the real number $L$ as its limit, if the distance of $a_n$ from $L$ 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 $N$ (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} |
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| \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.\\ |
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\textbf{Example.}\, Using the notation (2) we can write a result
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\textbf{Example.}\, Using the notation (2) we can write the result
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| $$\lim_{n\to\infty}\frac{n}{n\!+\!1} = 1.$$ |
$$\lim_{n\to\infty}\frac{n}{n\!+\!1} = 1.$$ |
| It's a question of that the real number sequence |
It's a question of that the real number sequence |
|
$$\frac{1}{2},\;\frac{2}{3},\;\frac{3}{4},\;\ldots$$
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$$\frac{1}{2},\,\frac{2}{3},\,\frac{3}{4},\,\ldots$$
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| has the limit value 1 (e.g. the nine hundred ninety-ninth member $\frac{999}{1000}$ is already ``almost'' 1!).\, For justificating the result, let $\varepsilon$ be an arbitrary positive number, as small as you want.\, Then |
has the limit value 1 (e.g. the nine hundred ninety-ninth member $\frac{999}{1000}$ is already ``almost'' 1!).\, For justificating the result, let $\varepsilon$ be an arbitrary positive number, as small as you want.\, Then |
| \begin{align} |
\begin{align} |
| \left|1-\frac{n}{n\!+\!1}\right| = \left|\frac{n\!+\!1}{n\!+\!1}-\frac{n}{n\!+\!1}\right| |
\left|1-\frac{n}{n\!+\!1}\right| = \left|\frac{n\!+\!1}{n\!+\!1}-\frac{n}{n\!+\!1}\right| |
| = \left|\frac{1}{n\!+\!1}\right| = \frac{1}{n\!+\!1} < \varepsilon, |
= \left|\frac{1}{n\!+\!1}\right| = \frac{1}{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{1}{\varepsilon}-1. |
n > \frac{1}{\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 $n$.\\ |
| $N = \frac{1}{\varepsilon}\!-\!1$.\\ |
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| [Not ready...] |
[Not ready...] |
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