+26404 $$\lim \limits_{n\rightarrow +\infty} {\sqrt{(n^2-n)}-n } =\ ?$$
Wir formen um:
$$=\lim \limits_{n\rightarrow +\infty} {\sqrt{n(n-1)}-n }\\
=\lim \limits_{n\rightarrow +\infty} {\sqrt{\frac{n}{n}\cdot n(n-1)}-n }\\
=\lim \limits_{n\rightarrow +\infty} {\sqrt{\frac{1}{n}\cdot n^2(n-1)}-n }\\
=\lim \limits_{n\rightarrow +\infty} { n \sqrt{\frac{n-1}{n}}-n }\\
=\lim \limits_{n\rightarrow +\infty} { n \sqrt{1-\frac{1}{n}}-n }\\
=\lim \limits_{n\rightarrow +\infty} { n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }\\$$
Die Taylorreihen-Entwicklung von 
mit Entwicklungspunkt 1
Wir setzen für $$\small{\text{$x=-\frac{1}{n}$}}$$
$$\small{\text{
$
\left(1-\frac{1}{n}\right)^{\frac{1}{2}}
=
1
+ \dfrac{1}{2}\cdot \left( -\dfrac{1}{n} \right)
- \dfrac{1}{8}\cdot \left( -\dfrac{1}{n} \right)^2
+ \dfrac{1}{16}\cdot \left( -\dfrac{1}{n} \right)^3
- \dfrac{5}{128}\cdot \left( -\dfrac{1}{n} \right)^4
\pm \cdots
$
}}\\\\
\small{\text{
$
=
1
- \dfrac{1}{2}\cdot \left( \dfrac{1}{n} \right)
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n} \right)^2
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n} \right)^3
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n} \right)^4
\pm \cdots
$
}}\\\\
\small{\text{$
{ n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }
=
n\cdot \left[
1
- \dfrac{1}{2}\cdot \left( \dfrac{1}{n} \right)
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n} \right)^2
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n} \right)^3
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n} \right)^4
\pm \cdots \right] -n
$}}\\\\
\small{\text{$
{ n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }
=
\not{n}
- \dfrac{1}{2}\cdot \left( \dfrac{n}{n} \right)
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n^1} \right)
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n^2} \right)
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n^3} \right)
\pm \cdots -\not{n}
$}}\\\\
\small{\text{$
{ n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }
=
- \dfrac{1}{2}\cdot \left( \dfrac{n}{n} \right)
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n^1} \right)
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n^2} \right)
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n^3} \right)
\pm \cdots
$}}\\\\
\small{\text{$
{ n \textcolor[rgb]{1,0,0}{\left(1-\frac{1}{n}\right)^{\frac{1}{2}} } -n }
=
- \dfrac{1}{2}
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n^1} \right)
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n^2} \right)
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n^3} \right)
\pm \cdots
$}}\\\\
\small{\text{
$
\lim \limits_{n\rightarrow +\infty} \left[ - \dfrac{1}{2}
- \dfrac{1}{8}\cdot \left( \dfrac{1}{n^1} \right)
- \dfrac{1}{16}\cdot \left( \dfrac{1}{n^2} \right)
- \dfrac{5}{128}\cdot \left( \dfrac{1}{n^3} \right)
\pm \cdots \right] = -\dfrac{1}{2}
$}}$$
+14538 $${\mathtt{U}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,\times\,}}{\mathtt{\pi}}$$ $${\mathtt{U}} = {\mathtt{d}}{\mathtt{\,\times\,}}{\mathtt{\pi}}$$
http://web2.0rechner.de/formelsammlung/mathematik/geometrie/kreis
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