first commit

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2026-03-11 07:08:36 +01:00
commit 90d3311e7d
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\begin{tikzpicture}
\usetikzlibrary{plotmarks}
\draw[color=gray, thin] (-3,0) -- (3,0);
%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
\draw[fill=blue!50,color=blue!50, ultra thick] (-1.5, 0) circle (.075);
\draw[fill=blue!50,color=blue!50, ultra thick] (1.5, 0) circle (.075);
\node[color=blue!50,] at (-1.5,0.35) {\textbf{a}};
\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
\end{tikzpicture}

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++++\begin{tikzpicture}
\usetikzlibrary{plotmarks}
\draw[color=gray, thick] (-3,0) -- (3,0);
%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
\node[color=blue!50] at (-1.5,0.35) {A};
\node[color=blue!50] at (1.5,0.35) {B};
\end{tikzpicture}

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\begin{tikzpicture}
\usetikzlibrary{plotmarks}
\draw[color=gray, thin] (-3,0) -- (3,0);
%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
\node[color=blue!50,] at (-1.5,0.35) {\textbf{a}};
\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
\end{tikzpicture}

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\begin{tikzpicture}
\usetikzlibrary{plotmarks}
\draw[color=gray, thin] (-3,0) -- (3,0);
%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
\draw[fill=blue!50,color=blue!50, ultra thick] (-1.5, 0) circle (.075);
\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
\node[color=blue!50] at (-1.5,0.35) {\textbf{a}};
\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
\end{tikzpicture}

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\begin{tikzpicture}
\usetikzlibrary{plotmarks}
\draw[color=gray, thin] (-3,0) -- (3,0);
%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
\draw[fill=blue!50,color=blue!50, ultra thick] (1.5, 0) circle (.075);
\node[color=blue!50] at (-1.5,0.35) {\textbf{a}};
\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
\end{tikzpicture}

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\begin{pspicture}(-1,-1)(4.9,4.9)
%\psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
%%%%%%%%%%%%%%%%%%%%%%%%Koor%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psaxes[Ox=0,Dx=2,Oy=0,Dy=2,linewidth=0.1pt]{->}(0,0)(0,0)(4.5,4.5)%Schnittpunkt, x0, y0 x1,y1
\rput[l](1.7,-0.8){$1.\ Koordinate$}
\rput[c](4.8,0){$x$}
\rput[l](0.4,4){$2.\ Koordinate$}
\rput[c](0,4.8){$y$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Vektor / Gerade%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline[linecolor=red, linewidth=0.1pt](0,0)(3,3)
\rput[c](3.0,-0.3){$a_1$}
\rput[c](-0.3,3){$a_2$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%Hlfslinien%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](3,-0.2)(3,3)
\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](-0.2,3)(3,3)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Pfeil%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline{->}(3.5,2.5)(3.1,2.9)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Kreuz %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline[linecolor=green] (2.9,3.1)(3.1,2.9)
\psline[linecolor=green] (2.9,2.9)(3.1,3.1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Beschriftung am Pfeil %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\rput[l](3.3,2.3){Punkt}
\rput[l](3.2,1.9){$(a_1, a_2)$}
\end{pspicture}

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\begin{pspicture}(-1,-1)(4.9,4.9)
%\psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Koordinaten%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psaxes[Ox=0,Dx=2,Oy=0,Dy=2,linewidth=0.1pt]{->}(0,0)(-0.0,-0.0)(4.5,4.5)%Schnittpunkt, x0, y0 x1,y1
\rput[c](4.8,0){$x$}
\rput[c](0,4.8){$y$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Vektor%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline[linecolor=red, linewidth=0.1pt]{->}(0,0)(3,3)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Hilfslinien%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](3,-0.2)(3,3)
\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](-0.2,3)(3,3)
\rput[c](3.0,-0.3){$a_1$}
\rput[c](-0.3,3){$a_2$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%Pfeil mit Beschriftung%%%%%%%%%%%%%%%%%%%%%%%%%%%
\psline{->}(3.5,2.5)(3.1,2.9)
\rput[l](1.8,2.3){$Vektor\ ("`Pfeil"')$}
\rput[l](2.5,1.9){$\vec{a}$}
\end{pspicture}

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definitions.tex Normal file
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%!TEX root=main.tex
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%a4paper,
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left=20mm,
top=20mm,
right=40mm,
marginparwidth=30mm
}
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%%
% Just some sample text
\usepackage{lipsum}
\usepackage{wrapfig}
%%
% For nicely typeset tabular material
\usepackage{booktabs}
%\usepackage[utf8]{inputenc}
%%
% For graphics / images
\usepackage{graphicx}
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\usepackage{mathtools}
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\usepackage{cmbright}
%%
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% up no horizontal space). Useful in tabular environments.
\newcommand{\hangp}[1]{\makebox[0pt][r]{(}#1\makebox[0pt][l]{)}}
%%
% Prints an asterisk that takes up no horizontal space.
% Useful in tabular environments.
\newcommand{\hangstar}{\makebox[0pt][l]{*}}
%%
% Prints a trailing space in a smart way.
\usepackage{xspace}
%\usepackage[makeroom]{cancel}
%%
% Some shortcuts for Tufte's book titles. The lowercase commands will
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% will print out the full title of the book in italics.
%\newcommand{\vdqi}{\textit{VDQI}\xspace}
%\newcommand{\ei}{\textit{EI}\xspace}
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%\newcommand{\VDQI}{\textit{The Visual Display of Quantitative Information}\xspace}
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\usepackage{cancel}
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\usetikzlibrary{fadings,shapes.arrows,shadows}
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%\usepackage{mathrsfs}
\usepackage{tabu}
\usepackage{textcomp}
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folgen01.tex Normal file
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%!TEX root=main.tex
\section{Folgen}
\paragraph*{Beispiel 1:}
Monatlicher Umsatz eines Shops: \\
\begin{center}
\begin{tabular}{|l|cccccc|}
\hline
\rule{0pt}{12pt}{\cellcolor[rgb]{1,0.647,0}}Monat &1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\rule{0pt}{12pt}{\cellcolor[rgb]{1,0.647,0}}Umsatz in T\texteuro& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tikzpicture}[scale=1.25]
\definecolor{orange1}{rgb}{1,0.647,0}
\begin{axis}[%
xlabel={Monat},
ylabel={Umsatz [T\texteuro]},
%clickable coords={(xy): \thisrow{label}},%
scatter/classes={%
a={mark=square*,orange1}}]
\addplot[scatter,only marks,%
scatter src=explicit symbolic]%
table[meta=label] {
x y label
1 1 a
2 2 a
3 3 a
4 4 a
5 5 a
6 6 a
};
\end{axis}
\end{tikzpicture}
\end{center}
\paragraph*{Beispiel 2:}
Rückzahlung eines Kredites (monatlich 200 \texteuro): \\
\begin{center}
\begin{tabular}{|l|cccccc|}
\hline
\rule{0pt}{12pt}{\cellcolor[rgb]{0.98,0.1,0}}Monat &1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\rule{0pt}{12pt}{\cellcolor[rgb]{0.98,0.1,0}}Schuld in T\texteuro& 7,0 & 6,8 & 6,6 & 6,4 & 6,2 & 6,0 \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tikzpicture}[scale=1.25]
\definecolor{red1}{rgb}{0.98,0.1,0}
\begin{axis}[%
xlabel={Monat},
ylabel={Schuld [T\texteuro]},
%clickable coords={(xy): \thisrow{label}},%
scatter/classes={%
a={mark=square*,red1}}]
\addplot[scatter,only marks,%
scatter src=explicit symbolic]%
table[meta=label] {
x y label
1 7.0 a
2 6.8 a
3 6.6 a
4 6.4 a
5 6.2 a
6 6.0 a
};
\end{axis}
\end{tikzpicture}
\end{center}
\fancybox[Definition \glqq undendliche Folge\grqq]{Eine unendliche Folge is eine Abbildung, die jeder natürlichen Zahl $n$ (also $n \in \mathbb{ N}$ bzw. $n \in \mathbb{N}_0$) eine reelle Zahl $a_n$ zuordnet.
1 - 4:52 weiter
}
\vfill
x-Achse = Abzisse
y-Achse = Ordinate
%\def\hcenter#1{\hfil#1\hfil}
%\begin{table}[]
% \begin{tabu}{
% >{\columncolor[HTML]{9A0000}}X[l]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|}
% {\color[HTML]{FFFFFF} \textbf{Monat}} & {\color[HTML]{9A0000} \textbf{1}} & {\color[HTML]{9A0000} \textbf{2}} & {\color[HTML]{9A0000} \textbf{3}} & {\color[HTML]{9A0000} \textbf{4}} & {\color[HTML]{9A0000} \textbf{5}} & {\color[HTML]{9A0000} \textbf{6}}\\
% \rule{10pt}{20pt}{\color[HTML]{FFFFFF} \textbf{Umsatz in T\texteuro}} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}
% \end{tabu}
%\end{table}
% Muster
%\begin{tabular}{|>{\columncolor{blue!40}}r|rrrrr|}
% \hline
% \rowcolor[gray]{.8} \textbf{No.} & {\bf 134} & {\bf 135} & {\bf 136} & {\bf 137} & {\bf 138} \\
% \hline
% \textbf{Milch } & 0.00 & 0.05 & 0.00 & 0.04 & 0.00 \\
% \textbf{Käse } & 49.57 & 49.15 & 49.80 & 49.68 & 50.18 \\
% \textbf{Zucker } & 0.01 & 0.00 & 0.00 & 0.00 & 0.00 \\
% {\bf Apfel } & 0.00 & 0.06 & 0.00 & 0.01 & 0.01 \\
% {\bf Wurst } & 46.14 & 46.56 & 46.32 & 46.48 & 46.31\\\hline
% {\bf Total } & {\bf 97.13} & {\bf 97.23} & {\bf 97.53} & {\bf 97.65} & {\bf 98.04} \\\hline
%\end{tabular}
\newpage

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set terminal png
set output "gnuplot01.png"
set grid
set samples 100000
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-7:10]
set xrange [-8:8]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set ytics axis
set xtics axis
#set object circle at 3,3 size 0.2
set label "(2,0.33)" at 2.1,0.77 tc rgb "#FF0000"
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle, '-' w p pt 7 ps 1.5 lc rgb "#FF0000" notitle
2.0 0.33
e
#"<echo '1 2'" with points pt 7 ps 2 #pt = filled circle ps = size

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set terminal png
set output "gnuplot01a.png"
set grid
unset border
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-7:10]
set xrange [-8:8]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set ytics axis
set xtics axis
set arrow from 3,-5 to -0.9,-5 lt rgb "#FF0000" lw 2
set label "von rechts" at 0.9,-4.5 tc rgb "#FF0000"
plot 1/(x**2 -1) lt rgb "#000FCF" lw 2 notitle

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set terminal png
set output "gnuplot01b.png"
set grid
set samples 100000
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-7:10]
set xrange [-8:8]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set ytics axis
set xtics axis
set arrow from -4.2,6 to -1.1,6 lt rgb "#FF0000" lw 2
set label "von links" at -3.5,6.5 tc rgb "#FF0000"
;set style line 1 lt 2 lw 2 pt 3 ps 0.5
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle

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set terminal png
set output "gnuplot01.png"
set grid
set samples 100000
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-7:10]
set xrange [-8:8]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set ytics axis
set xtics axis
#set object circle at 3,3 size 0.2
set label "(2,0.33)" at 2.1,0.77 tc rgb "#FF0000"
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle, '-' w p pt 7 ps 1.5 lc rgb "#FF0000" notitle
2.0 0.33
e
#"<echo '1 2'" with points pt 7 ps 2 #pt = filled circle ps = size

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\centering
\begin{gnuplot}[terminal=pdf,terminaloptions={font ",14" linewidth 2}]
set nogrid
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-1:10]
set xrange [-8:3]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
;set xtics axis out scale 0.5
set ytics axis
plot exp(x) lt rgb "#0000FF"
\end{gnuplot}

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set terminal png
set output "gnuplot02a.png"
set size 10./10.
set nogrid
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-8:8]
set xrange [-8:8]
set key center top reverse Left
set xzeroaxis
set yzeroaxis
;set xtics axis out scale 0.5
set ytics axis
plot 1/x lt rgb "#FF00FF" notitle

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set terminal png
set output "gnuplot01b.png"
set grid
set samples 100000
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-7:10]
set xrange [-8:8]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set ytics axis
set xtics axis
set arrow from -4.2,6 to -1.1,6 lt rgb "#FF0000" lw 2
set label "von links" at -3.5,6.5 tc rgb "#FF0000"
;set style line 1 lt 2 lw 2 pt 3 ps 0.5
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle

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\begin{gnuplot}[terminal=pdf,terminaloptions={font ",10" linewidth 2}]
unset border
set nogrid
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-3:5]
set xrange [-1:8]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot log(x) lt rgb "#FF0000"
\end{gnuplot}

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set terminal png
set output "gnuplot03a.png"
set nogrid
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-1:10]
set xrange [-8:3]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
;set xtics axis out scale 0.5
set ytics axis
plot exp(x) lt rgb "#0000FF" notitle

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\centering
\begin{gnuplot}[terminal=pdf,terminaloptions={font ",14" linewidth 2}]
set nogrid
unset border
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-1:10]
set xrange [-8:3]
set size 10./10.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
;set xtics axis out scale 0.5
set ytics axis
plot exp(x) lt rgb "#0000FF"
\end{gnuplot}

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\begin{gnuplot}[terminal=pdf,terminaloptions={font ",10" linewidth 2}]
unset border
set nogrid
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-12:-1]
set xrange [-4:4]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(sqrt(4-x)/(sqrt(12+x)-4)) lt rgb "#FFAA00"
\end{gnuplot}

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set terminal png
set output "gnuplot04a.png"
unset border
set nogrid
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-3:5]
set xrange [-1:8]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot log(x) lt rgb "#FF0000" notitle

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\begin{gnuplot}[terminal=pdf,terminaloptions={font ",10" linewidth 2}]
unset border
set nogrid
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-15:-4]
set xrange [-4:4]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(-x**2 +2*x -8) lt rgb "#FFAA00"
\end{gnuplot}

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set terminal png
set output "gnuplot05a.png"
unset border
set nogrid
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-15:-3]
set xrange [-4:4]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(-x**2 +2*x -8) lt rgb "#FFAA00" lw 2 notitle

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set terminal png
set output "gnuplot06a.png"
unset border
set nogrid
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-1:9]
set xrange [-10:10]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(sqrt(64-x**2)) lt rgb "#CBAA00" lw 2 notitle

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set terminal png
set output "gnuplot07a.png"
unset border
set nogrid
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-6:6]
set xrange [-10:30]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(7/(9-x)) lt rgb "#CBAAC0" lw 2 notitle

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set terminal png
set output "gnuplot08a.png"
unset border
set nogrid
set samples 100000
set lmargin at screen 0
set rmargin at screen 1
set bmargin at screen 0
set tmargin at screen 1
set yrange [-1:6]
set xrange [-1:20]
set size 5/5.
set key center top reverse Left
set xzeroaxis
set yzeroaxis
set xtics axis out scale 0.5
set ytics axis
plot(sin(x)**(1/log(x))) lt rgb "#CBAAC0" lw 2 notitle

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\begin{tikzpicture}[scale=1.5]
\pgfplotsset{compat=1.11}
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
single arrow head extend=.3cm,}}
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
%}
%\node [
% fill=blue!50, draw,
% single arrow, single arrow head indent=0ex,
% rotate=0,
% font=\sffamily
%] at (1,1.5)
%{\rotatebox{0}{ \qquad}};
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
\begin{axis}[
thin,
axis x line=center,
axis y line=center,
%xtick={-2,...,2},
ytick={-1,...,6},
grid = both,
minor x tick num=5,
minor y tick num=5,
%minor x tick style={line width=0pt},
major x grid style={black, line width=0.2pt},
major y grid style={black, line width=0.2pt},
minor x grid style={gray, line width=0.1pt},
%minor y tick num={5},
%minor y tick style={line width=0pt},
%yminorgrids,
%minor y grid style={black, line width=0.1pt},
xlabel style={below right},
ylabel style={above left},
minor tick length=0pt,
xmin=-2.9,
xmax=2.9,
ymin=-1.2,
style={>=latex},
yticklabel style = {font=\footnotesize,xshift=0.5ex},
xticklabel style = {font=\footnotesize,yshift=0.5ex},
% grid=both,
% grid style={line width=.1pt, draw=gray!10},
% major grid style={line width=.2pt,draw=gray!50},
ymax=5.5]
\addplot [mark=none,domain=-3.8:3.8, color=NewBlue, line width=0.75mm,smooth] {x^2};
\end{axis}
\draw[-{Triangle[scale=3,length=5,width=6]}, color=NewBlue, line width=3mm, fill=white] (2,1.5) to (0.8,1.5);
%http://latexcolor.com
\draw[-{Triangle[scale=3,length=5,width=6]}, color=NewBlue, line width=3mm] (4.8,1.5) to (6,1.5);
\draw[draw=NewBlue, fill=white, line width=0.75mm] (-1,1.2) rectangle (0.5,1.8) node[pos=.5] {\textbf{\textcolor{black}{$-\infty$}}};
\draw[draw=NewBlue, fill=white, line width=0.75mm] (6.3,1.2) rectangle (7.8,1.8) node[pos=.5] {\textbf{\textcolor{black}{$+\infty$}}};
%\tikzfancyarrow[3cm]{} arrow
\end{tikzpicture}

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\begin{tikzpicture}[scale=0.8]
\pgfplotsset{compat=1.11}
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
single arrow head extend=.3cm,}}
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
%}
%\node [
% fill=blue!50, draw,
% single arrow, single arrow head indent=0ex,
% rotate=0,
% font=\sffamily
%] at (1,1.5)
%{\rotatebox{0}{ \qquad}};
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
\begin{axis}[
x=1cm,y=1cm,
axis x line=center,
axis y line=center,
%axis lines=middle,
ymajorgrids=true,
xmajorgrids=true,
xmin=-5,
xmax=5,
ymin=-6,
ymax=6,
xtick={-5,-4,...,5},
ytick={-5,-4,...,6},]
\addplot [mark=none,domain=-4.8:-1.05, color=NewBlue, line width=0.5mm,step=10000, smooth, tension=0.2] {1/(x^2-1)};
\addplot [mark=none,domain=-0.95:0.95, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
\addplot [mark=none,domain=1.05:4.8, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
\end{axis}
\filldraw[red](7,6.33) circle (0.75mm) node[above,right, yshift=4]{(2,0.33)};
%\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
\end{tikzpicture}

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%!TEX root=main.tex
\section{Grenzwerte}
\subsection{Der Limes ~\cite{studimup.de}}
Mit dem Limes können Grenzwerte angegeben werden. Der Limes beschreibt, was passiert, wenn man für eine Variable Werte einsetzt, die einem bestimmten Wert immer näherkommen. Dabei steht unter dem „lim“ die Variable und gegen welche Zahl sie geht (also welchem Wert die Variable immer näher kommt). Nach dem „lim“ steht dann die Funktion, worin dann die Werte für x eingesetzt werden, zum Beispiel:
$\lim _ { x \rightarrow \infty } \frac { 1 } { x }$
Diese Schreibweise bedeutet, dass man für $x$ in die Funktion $1/x$ Werte einsetzt, die immer näher an unendlich herankommen. Man kann ja keinen unendlichen Wert einsetzen, aber man kann mit dem Limes „gucken“ was für unendlich herauskommen würde. Man spricht dann „Limes gegen unendlich“. Das geht natürlich auch mit allen anderen Werten, nicht nur für unendlich.
\subsection{Grenzwerte im Unendlichen}
Grenzwerte im Unendlichen beschreiben, was mit der Funktion passiert, also an welchen Wert sich die Funktion immer mehr annähert, wenn $x$ gegen unendlich läuft (das heißt, wenn $x$ immer größer wird bis unendlich). Dabei kann $x$ gegen $+\infty$ und $-\infty$ laufen, also immer kleiner oder größer werden. Es sieht dann in mathematischer Schreibweise folgendermaßen aus:
$\lim _ { x \rightarrow \infty } f ( x ) \quad$ und $\quad \lim _ { x \rightarrow - \infty } f ( x )$
\begin{center}
\input{limit001.pgf}
\end{center}
\subsection{Merkblatt}\label{lbl:MerkblattGrenzwert}
\subsubsection{Wichtige Grenzwerte}
% \setlength\extrarowheight{10pt}
%\begin{tabular}{|C{2cm}|C{2cm}|}
% \hline
% & \\
% \hline
% $\frac { 1 } { \infty }$ & $0$ \\
% \hline
% $\frac { 1 } { \pm 0 }$ & $\pm \infty$\\
% \hline
% & \\
% \hline
% & \\
% \hline
% & \\
% \hline
% & \\
% \hline
% & \\
% \hline
%\end{tabular}
%\begin{tabular}{l}
\begin{TAB}(r,7mm,8mm)[10pt]{|c|l|}{|c|c|c|c|c|c|}% (rows,min,max)[tabcolsep]{columns}{rows}
$\frac { 1 } { \infty }$ & $0$ \\
$\frac { 1 } { \pm 0 }$ & $\pm \infty$ \\
$\mathrm { e } ^ { \infty }$ & $\infty$\\
$q ^ { \infty }$ & $\text {falls } | q | < 1$\\
$\ln ( \infty )$&$\infty$\\
$\ln \left( 0 ^ { + } \right) $&$ - \infty$
\end{TAB}
\begin{longtable}{|l|l|l|}
$\left[ \frac { 0 } { 0 } \right] , \left[ \frac { \infty } { \infty } \right]$ & \textbf{Zähler} und \textbf{Nenner} einzeln \textbf{ableiten}. & $\lim _ { x \rightarrow 0 } \frac { \sin ( x ) } { x } \frac { \left[ \frac { 0 } { 0 } \right] } { = } \lim _ { x \rightarrow 0 } \frac { \cos ( x ) } { 1 } = 1$\\\hline
& & $\lim _ { x \rightarrow \infty } \frac { 3 x } { e ^ { 2 x } } \stackrel { \left[ \frac { \infty } { \infty } \right] } { = } \lim _ { x \rightarrow \infty } \frac { 3 } { 2 e ^ { 2 x } } = 0$ \\
\\
$[ 0 \cdot \infty ]$ & \textbf{Umformen zu} $\left[ \frac { 0 } { 0 } \right]$ \textbf{oder} $\left[ \frac { \infty } { \infty } \right]$: & \\
& \textbf{Bruch vorhanden?}: Bruch \glqq zusammensetzen\grqq\ & $\lim _ { x \rightarrow \infty } \frac { 1 } { x } \cdot \ln ( x ) \stackrel { [ 0 \cdot \infty ] } { = }$ \\
& & $\lim _ { x \to \infty } \frac { \ln ( x ) } { x } \stackrel { \left[ \frac { \infty } { \infty } \right] } { = } \ldots$ \\ \\
& \textbf{Sonst}: Bruch \glqq erzeugen\grqq mit $a \cdot b = \frac { a } { 1 / b } = \frac { b } { 1 / a }$& $\lim _ { x \rightarrow 0 ^ +} x \cdot \ln ( x ) \mathop =\limits^{\left[ 0 \cdot \infty \right]} $\\
& & $\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln ( x ) } { 1 / x } \mathop = \limits^{\left[ \frac{\infty }{\infty } \right]}\ldots$\\ \\
$ \left[ 0 ^ 0 \right] , \left[ 1 ^ \infty \right],$ & \textbf{Umformen zu} $\left[ 0 \cdot \infty \right] : \quad a ^ b = e ^ { b \cdot \ln ( a ) }$ & $\lim _ { x \rightarrow 0 ^ { + }
} ( 1 - x ) ^ { \frac { 1 } { x } } \stackrel { \left[ 1 ^ { \infty } \right] } { = }$\\
$\left[ \infty ^ 0 \right]$& & $\lim _ { x \rightarrow 0 ^ { + } } e ^ { \frac { 1 } { x } \cdot \ln ( 1 - x ) } \stackrel { [ 0 \cdot \infty ] } { = } \ldots$\\
$[ \infty - \infty ]$ & \textbf{Umformen}: &\\
&\textbf{Bruch vorhanden?}: Hauptnenner bilden & $\lim _ { x \rightarrow 0 ^ { + } } \frac { 1 } { x } - \frac { 1 } { \ln ( 1 + x ) } \stackrel { [ \infty - \infty ] } { = }$\\
& & $\stackrel { [ \infty - \infty ] } { = }\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln ( 1 + x ) - x } { x \cdot \ln ( 1 + x ) } \stackrel { \left[ \frac { 0 } { 0 } \right] } { = } \ldots$\\
& \textbf{Quadratwurzel vorhanden?:} $a - b = \frac { a ^ { 2 } - b ^ { 2 } } { a + b }$ & $\lim _ { x \rightarrow \infty } x - \sqrt { x ^ { 2 } + 1 } \stackrel { [ \infty - \infty ] } { = }$\\
\end{longtable}
\vfill
\pagebreak
\subsection{Wichtige Grenzwerte}
Empfehlung: Die Zahl die gegen das $x$ läuft in die Funktion einsetzen. Im einfachsten Fall kommt sofort das Endergebnis heraus.
\begin{enumerate}
\item Einfach
z.B. $\mathop {\lim }\limits_{x \to - 3} {x^2} = 9$
\item $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0$
\item $\mathop {\lim }\limits_{x \to {2^ + }} \frac{{\ln \left( {x - 1} \right)}}{{x - 2}} = 1$
\end{enumerate}
\textbf{Beispiel}: Funktion $\mathop {\lim }\limits_{x \to - 2} \frac{1}{{{x^2} - 1}}$
\begin{figure}[h]
% \pgfplotsset{compat=1.13}
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% \clip(-7.72,-5.48) rectangle (7.72,5.48);
% \draw[line width=0.75pt,color=qqwuqq,smooth,samples=400,domain=-7.72:7.719999999999998] plot(\x,{1/((\x)^(2)-1)});
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% %\draw[color=qqwuqq] (-7.54,-0.05) node {$f$};
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% \draw[color=ffqqqq] (2.8,0.77) node {$(2, 0.33)$};
% \end{scriptsize}
% \end{axis}
% \end{tikzpicture}
\centering
%\includegraphics[scale=.7]{gnuplot01.png}
\input{limit01.pgf}
\end{figure}
%\begin{figure}[h]
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%\end{figure}
\marginpar{bekommt man allgemein bei dieser Methode ein eindeutiges Ergebnis, so ist die Aufgabe gelöst}Hier kann man auch erst alle $x$ mit dem Grenzwert füllen und berechnen. Hier in diesem Fall ergibt sich ein eindeutiges Ergebnis: $\mathop {\lim }\limits_{x \to - 2} \frac{1}{{{x^2} - 1}} = \frac{1}{3}$
Schaut man sich die Funktion an der Stelle $-1$ an, so kann man hier von zwei Seiten diesen Wert untersuchen. Wenn man von links kommt so schreibt man $\mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{{x^2} - 1}}$. Hier geht Funktion gegen $+\infty$. Von rechts $\mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{{x^2} - 1}}$ geht die Funktion gegen $-\infty$.
\definecolor{ffqqqq}{rgb}{1,0,0}
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%%%%%%%%%%%%%%Test%%%%%%%%%%%%%%
%\begin{figure}[htb]
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\begin{figure}[htb]
\centering
\begin{minipage}[t]{.45\linewidth}
\centering
\input{limit01a.pgf}
\caption*{An der Stelle $-2 \rightarrow 2^+$}
\end{minipage}%
\hfill%
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[scale=0.45]{gnuplot01b.png}
\caption*{An der Stelle $-2 \rightarrow 2^-$}
\end{minipage}%
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pagebreak
\begin{itemize}
\item $\frac{1}{{ \pm 0}} \to \pm \infty $ hiermit ist gemeint, je mehr der Nenner gegen $0$ geht, da eine Division durch Null nicht möglich ist.
\item $\frac{1}{\pm \infty} \to 0$
\end{itemize}
\begin{figure}[h]
\centering
%\resizebox{!}{.15\paperheight}{\input{gnuplot02.gp}}
%\resizebox{!}{.15\paperheight}{\includegraphics{gnuplot01a.png}}
\includegraphics[scale=0.5]{gnuplot02a.png}
\caption*{$\frac{1}{x}$}
\end{figure}
\begin{itemize}
\item $e^{\infty} \to \infty $
\item $e^{- \infty} \to 0$
\end{itemize}
\begin{figure}[h]
\centering
%TODO
%\resizebox{!}{.15\paperheight}{\input{gnuplot03a.gp}}
\includegraphics[scale=0.5]{gnuplot03a.png}
\caption*{$e^x$}
\end{figure}
\pagebreak
%http://tutorial.math.lamar.edu/Classes/CalcI/InverseFunctions.aspx
\marginpar{die Logarithmus-Funktion ist nur im positiven Bereich. Man kann sich nur von rechts der $0$ annähern.}
\begin{itemize}
\item $\ln (\infty) \to \infty $
\item $\ln (0^+) \to -\infty$
\end{itemize}
\begin{figure}[ht]
\centering
%\resizebox{!}{.15\paperheight}{\input{gnuplot03.gp}}
\includegraphics[scale=0.5]{gnuplot04a.png}
\caption*{$\ln x$}%
\end{figure}
\begin{itemize}
\item $q^\infty = 0$, falls $\left|q\right|<1$
\item $\ln (0^+) \to -\infty$
\end{itemize}
\vfill
\pagebreak
\subsection{Regeln von L'Hospital}
Wenn man Grenzwerte wie $\lim _ { x \rightarrow x _ { 0 } } f ( x )$ bestimmen soll, schaut man zuerst einmal, was man denn erhalten würde, wenn man $x_0$ einfach einsetzt.
Kommt bei der Berechnung von Grenzwerten einer der nachfolgenden Sonderfälle heraus, so kann man diese gegebenenfalls mit der Regeln von L'Hospital lösen.
\begin{description}
\item [1.]\tikz[na]\node [coordinate] (n1) {}; $\frac{0}{0}$
\item [2.]\tikz[na]\node [coordinate] (n2) {}; $\frac{\infty}{\infty}$
\end{description}
Diese beiden Punkte können mit der Regel von L'Hospital gelöst werden.
\textbf{Beispiel:} $\mathop {\lim }\limits_{x \to 2} {\frac{x^2+x-6}{x^2-3x+2}}$ wird hier nun die $2$ eingesetzt, so ergibt sich:
$\mathop {\lim }\limits_{x \to 2} {\frac{2^2+2-6}{2^2-3\cdot 2+2}} = \mathop {\lim }\limits_{x \to 2} {\frac{4+2-6}{4-6+2}} = \frac{0}{0}$. Da das Ergebnis $\frac{0}{0}$ beträgt wird nun die Regel von L'Hospital angewandt. Hierzu werden der Zähler und der Nenner separat differenziert. Somit ergibt sich $\mathop {\lim }\limits_{x \to 2} {\frac{2x+1}{2x-3}}=\frac{5}{1}=5$.
\begin{description}
\item [3.]\tikz[na]\node [coordinate] (n3) {}; $0\cdot \infty$ \qquad \tikz[na]\node [coordinate] (n31){}; Bruch vorhanden: $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( x \right)=\mathop {\lim }\limits_{x \to \infty } \frac{\ln \left( x \right)}{x}$
\hspace{1.35cm} \tikz[na]\node [coordinate] (n32){}; sonst: $\mathop {\lim }\limits_{x \to {0^ + }} x \cdot \ln \left( x \right)=\mathop {\lim }\limits_{x \to {0^ + }} \frac{\ln \left( x \right)}{\frac{1}{x}}$
%\begin{description}
%\item[]
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%\end{description}
\end{description}
\begin{description}
\item [4.]\tikz[na]\node [coordinate] (n41) {};$0^0$\tikz[na]\node [coordinate] (n4) {};
\item [5.]\tikz[na]\node [coordinate] (n51) {};$\infty^0$\tikz[na]\node [coordinate] (n5) {};
%$\mathop {\lim }\limits_{x \to {0^ + }} x \cdot \ln \left( x \right)=\mathop {\lim }\limits_{x \to {0^ + }} \frac{\ln \left( x \right)}{\frac{1}{x}}$
\item [6.]\tikz[na]\node [coordinate] (n61) {};$1^\infty$\tikz[na]\node [coordinate] (n6) {};
\item [7.]\tikz[na]\node [coordinate] (n7){};$\infty-\infty$ Bruch vorhanden: Hauptnenner bilden
$\sqrt[2]{\ldots} a-b=\frac{a^2-b^2}{a+b} $
$\sqrt[3]{\ldots} a-b =\frac{a^3-b^3}{a^2+ab+b^2}$
\end{description}
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\pagebreak
\subsubsection{Beispiele zu $\frac{0}{0}$ und $\frac{\infty}{\infty}$}
\begin{enumerate}
\item $\mathop {\lim }\limits_{x \to 2} {\frac{x^3-6x^2+12x-8}{x^2-4x+4}}=\frac{0}{0}$ Anwenden der Regel von L'Hospital.
$\mathop {\lim }\limits_{x \to 2} {\frac{3x^2-12x+12}{2x-4}}=\frac{0}{0}$ somit muss hier die L'Hospitalsche Regel noch einmal angewandt werden
$\mathop {\lim }\limits_{x \to 2} {\frac{6x-12}{2}}=\frac{0}{2} = \underline{\underline{0}}$
\item $\mathop {\lim }\limits_{x \to \infty}\frac{-2x^3+3x-1}{4\sqrt{5}+x^2+1}$
\marginpar{In der Unendlichkeit überleben immer nur die stärksten Terme}
Ausklammern der höchsten (des am stärksten wachsenden Term) Potenz in Nenner und Zähler. Die beiden höchsten Potenzen sind hier $x^3$ und $x^{\frac{5}{2}}$. Die am stärksten wachsende ist in diesem Fall $x^3$.
Somit $\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\frac{3}{x^2}-\frac{1}{x^3}\right)}{{x^3}\left( {4{x^{ - \frac{1}{2}}} + \frac{1}{x} + \frac{1}{{{x^3}}}} \right)}=\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\frac{3}{x^2}-\frac{1}{x^3}\right)}{{x^3}\left( {{\frac{4}{{{x^{\frac{1}{2}}}}}} + \frac{1}{x} + \frac{1}{{{x^3}}}} \right)} $\marginpar{$\frac{a^m}{a^n}=a^{m-n}$}
Alle Terme in der eine Konstante durch $x$ geteilt wird gehen nach $0$. Somit
$\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\cccancelto[red]{\frac{3}{x^2}}{0}-\cccancelto[red]{\frac{1}{x^3}}{0}\right)}{{x^3}\left( \bccancelto[orange]{\frac{ 4 }{x^{\frac{1}{2}}}}{0} + \bccancelto[orange]{\frac{1}{x}}{0} + \bccancelto[orange]{\frac{1}{x^3}}{0} \right)}$
Als Ergebnis erhält man hier $\frac{-2}{0}$ was in diesem Fall bedeutet das das Ergebnis nach unendlich geht, hier ist es $-\infty$
\item $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {{x^3}} \right)}}{{4{x^6}}}$ wird die $0$ eingesetzt so ergibt sich ein $\frac{0}{0}$ Ergebnis:
$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {{0^3}} \right)}}{{4 \cdot {0^6}}} = \frac{{1 - 1}}{0} = \frac{0}{0}$
Anwendung von L'Hospital: \marginpar{Kettenregel:
$f\left( x \right) = g\left( {h\left( x \right)} \right) \to f'\left( x \right) = g'\left(h\left( x \right) \right) \cdot h'\left( x \right)$ }
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^3}} \right) \cdot \cccancelto[blue]{3}{}{\cccancelto[red]{x^2}{}}}}{\bccancelto[blue]{24}{8} \cdot {x^{\bccancelto[red]{5}{3}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^3}} \right)}}{{8{x^3}}}$
Dies ergibt wieder $\frac{0}{0} \rightarrow \sin(0) = 0$ und $8\cdot 0^3 = 0$ somit wird wieder differenziert:
$\mathop {\lim }\limits_{x \to 0} \frac{\cos \left( x^3 \right) \cdot \cccancelto[green!50!black]{3}{}\cccancelto[orange]{x^2}{}}{\bccancelto[green!50!black]{24}{8}\bccancelto[orange]{x^2}{}}=\frac{1}{8}$
\item \marginpar{\[\frac{0}{{\sqrt {16} - 4}} = \frac{0}{0}\]}$\lim \limits_{x \to 4^-}\frac{\sqrt{4-x}}{\sqrt{12+x}-4} = \lim \limits_{x \to 4^-} \frac{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}{{{{\left( {12 + x} \right)}^{\frac{1}{2}}} - 4}}$
Nach dieser Umstellung kann man die L'Hospitalsche Regel anwenden:
Somit: $\mathop {\lim }\limits_{x \to 4^-}\frac{{\frac{1}{2}{{\left( {4 - x} \right)}^{ - \frac{1}{2}}} \cdot - 1}}{{\frac{1}{2}{{\left( {12 + x} \right)}^{ - \frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{ - \frac{1}{2}{{\left( {4 - x} \right)}^{ - \frac{1}{2}}}}}{{\frac{1}{2}{{\left( {12 + x} \right)}^{ - \frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{\frac{1}{2}{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{ - \frac{1}{2}{{\left( {4 - x} \right)}^{\frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{\cccancelto[black]{\frac{1}{2}}{}{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{ - \bccancelto[black]{\frac{1}{2}}{}{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}$
$ = - \frac{{{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}} = \frac{-4}{0}$
Die Division durch $0$ bedeutet in diesem Fall das das Ergebnis gegen unendlich geht, hier $-\infty$.
\item $\mathop {\lim }\limits_{x \to 0}\frac{e^x-2x-e^{-x}}{x-\sin{x}}$
\marginpar{Man kann die Regel solange anwenden, solange sich Änderungen ergeben, wen nicht sollte man ausklammern etc... ausprobieren}Einsetzen der $0$ ergibt
$\frac{{{e^0} - 2 \cdot 0 - {e^{ - 0}}}}{{0 - 0}} = \frac{0}{0}$
Somit muss wieder L'Hospital angewandt werden:
Ableiten von Nenner und Zähler: $\mathop {\lim }\limits_{x \to 0}\frac{e^x-2+e^{-x}}{1-\cos{x}}$
Einsetzen von $0$ ergibt: $\frac{{{e^0} - 2 + {e^{ - 0}}}}{{1 - \cos (0)}} = \frac{{1 - 2 + 1}}{{1 - 1}} = \frac{0}{0}$
Da noch immer keine Lösung erhalten wurde, wird nun wieder differenziert:
$\mathop {\lim }\limits_{x \to 0}\frac{e^x-2+e^{-x}}{1-\cos{x}}=\mathop{\lim }\limits_{x \to 0}\frac{e^x - e^{ - x}}{\sin (x)}$
Und wieder:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos (x)}} = \frac{2}{1} = 2$
\item $\mathop {\lim }\limits_{x \to \infty } \frac{{{2^x} - {2^{ - x}}}}{{{2^x} + {2^{ - x}}}}$
Hier wird ist nun ein Fall bei der die Regel von L'Hospital zwar zur Anwendung kommen kann, jedoch nichts bewirkt.
$\mathop {\lim }\limits_{x \to \infty } \frac{2^x - 2^{ - x}}{2^x + 2^{ - x}}=\mathop { \lim }\limits_{ x\to \infty }\frac { 2^{ x }\ln { ( } 2)-\left( -\ln { ( } 2)\cdot 2^{ -x } \right) }{ { 2 }^x\ln(2)-2^{-x}\ln(2) } =\mathop { \lim }\limits_{ x\to \infty }\frac { 2^{ x }\ln { \left(2\right)} +\ln { \left(2\right)}\cdot 2^{ -x } }{ { 2 }^{ x }\ln { \left(2\right)}-2^{ -x }\ln { \left(2\right)} } \\=\mathop {\lim }\limits_{ x\to \infty }\frac { \ln { \left( 2 \right) } \left( 2^{ x }+2^{ -x } \right) }{ \ln { \left( 2 \right) } \left( { 2 }^{ x }-2^{ -x } \right) } =\mathop {\lim }\limits_{ x\to \infty }\frac { \left( 2^{ x }+2^{ -x } \right) }{\left( { 2 }^{ x }-2^{ -x } \right) } $
Beim nächsten Differenzieren würde sich wiederum nur das Operationszeichen umkehren, also + wird wieder - etc. In diesem Fall wird ausgeklammert:
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\cccancelto[red]{\ln (2)}{}\left( {{2^x} + {2^{ - x}}} \right)}}{{\bccancelto[red]{\ln (2)}{}\left( {{2^x} - {2^{ - x}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{2^x} + {2^{ - x}}} \right)}}{{\left( {{2^x} - {2^{ - x}}} \right)}} $
Nun wird $2^x$ ausgeklammert.
Es gilt $\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \Rightarrow \frac{{{2^{ - x}}}}{{{2^x}}} = {2^{ - x - x}} = {2^{ - 2x}}$
und somit:
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{{2^x}\left( {1 + {2^{ - 2x}}} \right)}}{{{2^x}\left( {1 - {2^{ - 2x}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{1 + \cccancelto[blue]{2^{ - 2x}}{0}}{1 - \bccancelto[blue]{2^{ - 2x}}{0}}=\frac{1}{1}=1$
\end{enumerate}
\pagebreak
%}\frac{\left(4-x\right)^{-\frac{1}{2}}}{\left(12+x\right)^{-\frac{1}{2}}}} = \lim \limits_{x \to 4^-}\frac{{{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}$ \marginpar{wieder Anwendung der Kettenregel}
%fsdfsdfsdfsd
%\marginpar{
%\begin{wrapfigure}{l}{}
%\centering
%\resizebox{!}{.99\marginparwidth}{\input{gnuplot04.gp}}
%\caption{$\frac{\sqrt{4-x}}{\sqrt{12+x}-4}$}%
%\end{wrapfigure}
%}
%$\left\{
%\begin{tabular}{p{.8\textwidth}}
%\begin{itemize}
%\item Second line
%\item Third line, which is quite long and seemingly tedious in the extreme
%\item Fourth line, which isn't as long as the third
%\end{itemize}
%\end{tabular}
%\right.$
%$\xlimes{ 2 }{ \frac{ x^2 }{ x } }$
\subsubsection{Beispiele zu $0\cdot\infty$}
\begin{enumerate}
\item $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {2x} \right)$
Hier tritt beim Einsetzen von $\infty$ geht $\frac{1}{\infty}$ gegen $0$ und $\ln\left(2x\right)$ gegen $\infty$.
Umwandeln der Funktion zu einem Bruch, damit man den Fall $\frac{\infty}{\infty}$ bzw. $\frac{0}{0}$ erhält und somit die Regel von L'Hospital anwenden kann. Somit
\marginpar{$f(x)=\ln{x}$
$f'(x)=\frac{1}{x}$}
$\mathop {\lim }\limits_{x \to \infty } \frac{\ln \left( 2x\right)}{x}\Rightarrow \frac{\infty}{\infty}\Rightarrow \mathop {\lim }\limits_{x\to \infty }\frac { \frac { 1 }{ \cccancelto[red]{2}{}x } \cccancelto[red]{2}{} }{ 1 } = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0 $
\item $\mathop {\lim }\limits_{ x\to { 0 }^{ + } }x\cdot \ln { \left( 2x \right) } $
Hier ist $x=0$ und $\ln{\left(2\cdot 0^+\right)} = -\infty$ \marginpar{siehe wichtige Grenzwerte\ref{lbl:MerkblattGrenzwert}}
$\mathop {\lim }\limits_{ x\to { 0 }^{ + } }\frac { \ln { \left( 2x \right) } }{ \frac { 1 }{ x } } = \frac{\infty}{\infty} \Rightarrow $\marginpar{$\frac{1}{x}=x^{-1}=-x^{-2}=-\frac{1}{x2}$} $\frac{\frac{1}{\cccancelto[red]{2}{}x}\cccancelto[red]{2}{}}{-\frac{1}{x^2}}=\frac{x^2}{-x}=x=0$
\marginpar{$\frac{1}{\frac{1}{x}}=\frac{1}{x^{-1}}$}
\item $\mathop {\lim }\limits_{x\to\infty}X^3\cdot e^{-2x}\Rightarrow $ \marginpar{$\infty^3=\infty$ und $e^{-2\infty}=0$}
Da: $e^{-2x}=\frac{1}{e^{2x}}\Rightarrow \mathop {\lim }\limits_{x\to\infty}\frac{x^3}{e^{2x}}=\mathop {\lim }\limits_{x\to\infty}\frac{3x^2}{e^{2x}}\Rightarrow \frac{\infty}{\infty}\Rightarrow \mathop {\lim }\limits_{x\to\infty}\frac{6x}{2e^{2x}}= \mathop {\lim }\limits_{x\to\infty}\frac{6}{8e^{2x}}=0$ \marginpar{$\frac{6}{8e^{2\cdot\infty}}=0$}
\end{enumerate}
\subsubsection{Beispiele zu $0^0$, $\infty^0$ und $1^\infty$}
\begin{description}
\item[1.] $\mathop {\lim }\limits_{x\to 0^{+}} \tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n11){}; x^{-3 \tikz[na]\node [coordinate,yshift=1mm] (n21) {};x}=$ $0^{-3\cdot 0}\Rightarrow 0^0$ Umformen durch das Hinzufügen der Eulerschen Zahl $e$ als Basis, der alte Exponent hier $-\tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n22) {}; 3x$ bleibt und wird um den natürlichen Logarithmus der eigentlichen Basis ($\tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n12) {}; x$) erweitert:
$\mathop {\lim }\limits_{x\to 0^+}e^{-\tikz[na]\node [coordinate, xshift=1mm,yshift=-1mm] (n23) {};3x\ln\left(\tikz[na]\node [coordinate,yshift=-1mm] (n13) {}; x\right)}$
Jetzt nur der Exponent betrachtet:
\begin{equation*}\mathop{\lim }\limits_{ x\to 0^+ } x\cdot \left(\ln\left(\tikz[na]\node [coordinate,yshift=-2mm, xshift=-1mm] (n31) {};x\right)\right)\Rightarrow 0 \cdot \infty \Rightarrow \mathop{\lim }\limits_{ x\to 0^+ } \frac{\ln\left(x\right)}{\frac{1}{x}} \mathop \Rightarrow \limits^{\frac{\infty }{\infty }} \mathop = \limits^{L'H} \mathop{\lim }\limits_{ x\to 0^+ }\frac{\frac{1}{x}}{-\frac{1}{x^2}}=-\frac{x^{\cccancelto[red]{2}{}}}{\cccancelto[red]{x}{}}=x= {\color{red}\tikz[na]\node [coordinate, xshift=1mm,yshift=2mm] (n32) {};0}
\end{equation*}%\tikz[baseline, remember picture ]{\node[fill=blue!20,anchor=base] (t1) {$0$};}
\vspace{5mm}
Nun wird dieses Ergebnis in die Ausgangs-Exponentialgleichung eingesetzt:
\begin{equation*}
\mathop {\lim }\limits_{x \to {0^ + }} e^{-3 {\color{red}\ln\tikz[na]\node [coordinate, yshift=1.5mm] (n33) {};\left(x\right)}} = e^{-3 \cdot {\color{red}\tikz[na]\node [coordinate, yshift=1mm] (n34) {};0}} = e^0 = 1
\end{equation*}
\item[2.] $\mathop {\lim }\limits_{x \to 0 + } \sin {\left( x \right)^{\frac{1}{{\ln \left( x \right)}}}}\Rightarrow \sin
\left(
0
\right)=0 \text{ und } \frac{1}{\ln
\left(
0^+
\right)}=\frac{1}{-\infty}=0$
$\mathop {\lim }\limits_{x \to 0 + } e^{\frac{1}{\ln\left(x\right)}\cdot \ln\left(\sin\left(x\right)\right)}=\mathop {\lim }\limits_{x \to 0 + } e^{\frac{\ln
\left(
\sin
\left(
x
\right)
\right)}{\ln
\left(
x
\right)}}$
\marginpar{Man kann einzelne Teile aus dem kritischen Fall $\frac{0}{0}$ rauslösen, wenn man erstens nichts verändert und man zweitens keinen neuen kritischen Fall erzeugt. Da hier der $\cos\left(0\right)=1$ ist, kann dieser rausgelöst werden. }
\textbf{Nebenrechnung}:
$\mathop {\lim }\limits_{x \to 0 + }\frac{\ln
\left(
\sin
\left(
x
\right)
\right)}{\ln
\left(
x
\right)}=\mathop {\lim }\limits_{x \to 0 + }\frac{\frac{1}{\sin
\left(
x
\right)}\cdot\cos
\left(
x
\right)}{\frac{1}{x}}=\frac{\frac{\cos
\left(
x
\right)}{\sin
\left(
x
\right)}}{\frac{1}{x}}=\frac{x\cdot \cos\left(x\right)}{\sin\left(x\right)}$
$\underbrace {\mathop {\lim }\limits_{x \to 0 + }\cos
\left(
x
\right)}_{=1}\cdot\mathop {\lim }\limits_{x \to 0 + }\frac{x}{\sin
\left(
x
\right)}\mathop = \limits^{\frac{0}{0}}=\mathop {\lim }\limits_{x \to 0 + }\frac{1}{\cos
\left(
x
\right)}=1$
Dieses Ergebnis wird nun wieder in die umgeformte Ausgangsgleichung eingesetzt:
$\mathop {\lim }\limits_{x \to 0 + } e^{\frac{\ln
\left(
\sin
\left(
x
\right)
\right)}{\ln
\left(
x
\right)}}=e^1=\underline{\underline{e}}$
\end{description}
\begin{tikzpicture}[overlay]
\path[-latex, color=blue!40, line width=0.5mm, opacity=0.5] (n11) edge [bend left=70] (n12);
\path[-latex, color=red!40, line width=0.5mm, opacity=0.5] (n21) edge [bend left=70] (n22);
\path[-latex, color=red!40, line width=0.5mm, opacity=0.5] (n22) edge [bend left=70] (n23);
\path[-latex, color=blue!40, line width=0.5mm, opacity=0.5] (n12) edge [bend left=70] (n13);
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\newpage
\begin{description}
\item[3.] $\mathop {\lim }\limits_{x \to 1} \sqrt[{1 - x}]{x}=\mathop {\lim }\limits_{x \to 1}x^{\frac{1}{1-x}}=\mathop {\lim }\limits_{x \to 1}e^{\frac{1}{1-x}\cdot \ln\left(x\right)}=e^{-\tikz[na]\node [coordinate, yshift=1mm] (n42) {};1}$
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to 1}\frac{\ln
\left(
x
\right)}{1-x}= \mathop = \limits^{\frac{\infty }{\infty }} = \mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{x}}{-1}=-\tikz[na]\node [coordinate, yshift=1.5mm] (n41) {};1$
\item[4.]\marginpar{\raggedright{\color{red}$\frac{1}{x} = {x^{ - 1}} \left( {{x^{ - 1}}} \right)' = - {x^{ - 2}} = - \frac{1}{{{x^2}}}$
\color{blue}$1+\frac{a}{x}=\left(1+ax^{-1}\right)'=-ax^{-2}=-\frac{a}{x^2} $}
}
$\mathop {\lim }\limits_{x \to \infty}\left(1+\frac{a}{x}\right)^x =\mathop {\lim }\limits_{x \to \infty}e^{x\cdot\ln\left(1+\frac{a}{x}\right)}=e^{\tikz[na]\node [coordinate, xshift=1mm, yshift=1mm] (n52){};a}$
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to \infty}x\cdot\ln
\left(
1+\frac{a}{x}
\right)=\mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{a}{x}}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{1 + \frac{a}{}}} \cdot \left( { - \frac{a}{{{x^2}}}} \right)}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \frac{a}{{{x^2} \cdot 1 + \frac{a}{x}}}}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } - \frac{a}{\cccancelto[red]{x^2}{} \cdot 1 + \frac{a}{x}} \cdot -\cccancelto[red]{x^2}{} = \mathop {\lim }\limits_{x \to \infty } \frac{a}{{1 + \frac{a}{x}}}$ Einsetzen von $\infty$ in $x$ $\frac{a}{{1 + \frac{a}{\infty }}} = \frac{a}{{1 + 0}} = \tikz[na]\node [coordinate, yshift=1.5mm,xshift=1mm] (n51){};a$
\item[5.] $\mathop {\lim }\limits_{x \to \infty }
\left(10+x\right)^{\frac{2}{x}}=\mathop {\lim }\limits_{x \to \infty } e^{\frac{2}{x}\cdot\ln\left(10+x\right)}=e^0=1$
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to \infty }\frac{2\cdot
\ln\left(
10+x
\right)}{x}=\frac{2\cdot\frac{1}{10+x}\cdot 1}{1}=\frac{\frac{2}{10+x}}{1}=\frac{2\cdot\frac{1}{\infty}}{1}=\frac{0}{1}=0$
\item[6.]$\mathop {\lim }\limits_{x \to \infty } \left( e^{5x}-4x \right)^{\frac{1}{x}}=\mathop {\lim }\limits_{x \to \infty }e^{\frac{1}{x}\ln\left(e^{5x}-4x\right)} = e^{\textbf{\textcolor{orange}{5}}}$
Nebenrechnung: $\mathop {\lim }\limits_{x \to \infty }\frac{1}{x}\ln\left(e^{5x}-4x\right)=\mathop {\lim }\limits_{x \to \infty }\frac{\ln\left(e^{5x}-4x\right)}{x}=\mathop {\lim }\limits_{x \to \infty }\frac{1}{e^{5x}-4x}\cdot 5e^{5x}-4=\mathop {\lim }\limits_{x \to \infty}\frac{5e^{5x}-4}{e^{5x}-4x}=\mathop {\lim }\limits_{x \to \infty}{\frac{25e^{5x}}{5e^{5x}-4}}=\frac{125e^{5x}}{25e^{5x}}=\textbf{\textcolor{orange}{5}}$
\end{description}
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\path[-latex, color=green!40!black, line width=0.5mm, opacity=0.5] (n41) edge [bend left=-70] (n42);
\path[-latex, color=yellow!40!black, line width=0.5mm, opacity=0.5] (n51) edge [bend left=-50] (n52);
\end{tikzpicture}
\subsubsection{Beispiele zu $\infty - \infty$}
\begin{enumerate}
\marginpar{Im ersten Term wird im Zähler die 0 postiv, da $1^+$ und somit geht der Term gegen $\infty$}\item $\mathop {\lim }\limits_{x \to 1^+}\frac{1}{x-1}-\frac{1}{\ln\left(x\right)}$
Bilden eines Hauptnenners: $\frac{\ln\left(x\right)-\left(x-1\right)}{\left(x-1\right)\ln\left(x\right)}$ Dadurch entsteht der Fall $\frac{0}{0}$
\marginpar{$\left(u\cdot v\right)'=u'\cdot v+u\cdot v'$}$\mathop {\lim }\limits_{x \to 1^+}\frac{\ln\left(x\right)-\left(x-1\right)}{\left(x-1\right)\ln\left(x\right)}=\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{1}{x}-1}{1\cdot\ln\left(x\right)+\left(x-1\right)\cdot\frac{1}{x}}=\mathop {\lim }\limits_{x \to 1^+}\frac{-\frac{1}{x^2}}{\frac{1}{x} + \left( { - \frac{1}{{{x^2}}}} \right)\left( {x - 1} \right) + \frac{1}{x}}$
Einsetzen der $1$
$\frac{-1}{1-1\cdot 0 +1}=-\frac{1}{2}$
\marginpar{Über dritte binomische Formel: $a-b=\frac{a^2-b^2}{a+b}$ also $\left(a-b\right)\left(a+b\right)=a^2-b^2$}
\item $\mathop {\lim }\limits_{x \to \infty} x-\sqrt{x^2+1}=\mathop {\lim }\limits_{x \to \infty}\underbrace x_{\textcolor{red}{a}} - \underbrace {\sqrt {{x^2} - 1} }_{\textcolor{red}{b}}=\mathop {\lim }\limits_{x \to \infty}\frac{x^2-\left(x^2+1\right)}{x+\sqrt{x^2+1}}=\mathop {\lim }\limits_{x \to \infty}\frac{x^2-\left(x^2+1\right)}{x+\sqrt{x^2-1}}=\mathop {\lim }\limits_{x \to \infty}\frac{-1}{x+\sqrt{x^2-1}}=-\frac{1}{\infty}=0$
\item \marginpar{Zum Auflösen der Wurzel $a - b = \frac{{{a^3} - {b^3}}}{{{a^2} + ab + {b^3}}}$}$\mathop {\lim }\limits_{x \to \infty } x - \sqrt[3]{{{x^3} - {x^2}}}=\mathop {\lim }\limits_{x \to \infty } \underbrace x_a - \underbrace {\sqrt[3]{{{x^3} - {x^2}}}}_b=\mathop {\lim }\limits_{x \to \infty }\frac{{{x^3} - {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^3}}}{{{x^2} + x\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right) + {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}} =\mathop {\lim }\limits_{x \to \infty }\frac{{{x^3} - \left( {{x^3} - {x^2}} \right)}}{{{x^2} + x\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right) + {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}}$
Hier muss nun der am stärksten wachsende Term gesucht werden:
Zweiter Term des Nenners: $x\left( \sqrt[3]{x^3 - x^2} \right)$ hier würde beim Einsetzen von $\infty$ der Term $x\left( \sqrt[3]{x^3} \right)$ übrig, welcher sich dann wiederum auf $x\cdot x$ also $x^2$ reduziert.
Der dritte Term ${{{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}$ reduziert sich bei Betrachtung in der Unendlichkeit auf $\left( \sqrt[3]{x^3} \right)^2$. Da die dritte Wurzel aus $x^3 = x$ ist, bleibt $x^2$ übrig. Somit wachsen alle Terme im Nenner so schnell wie $x^2$.
Im Zähler bleibt da $x^3 - \left( x^3 - x^2 \right)$ nur $-x^2$ übrig, so also auch hier wächst alles so schnell wie $x^2$.
Das bedeutet das man jetzt den Faktor $x^2$ ausklammert:
\begin{itemize}
\item Term 2 des Nenners:
$\frac{x\left(\sqrt[3]{x^3 - x^2} \right)}{x^2} = \frac{\sqrt[3]{x^3 - x^2}}{x}$
Um nun die Nenner $x$ in die Wurzel zu bekommen muss dieser hoch 3 genommen werden, somit ergibt sich $\frac{\sqrt[3]{x^3 - x^2}}{x^3} = \sqrt[3]{\frac{x^3 - x^2}{x^3}}$
Nebenrechnung: $\frac{x^3 - x^2}{x^3} = \frac{x^2\left( x - 1 \right)}{x^2 \cdot x} = \frac{x - 1}{x} = 1 - \frac{1}{x}$
$=\sqrt[3]{1 - \frac{1}{x}}$
\item Term 3 des Nenners: $\frac{\left( \sqrt[3]{x^3 - x^2} \right)^2}{x^2} = {\left( {\frac{{\sqrt[3]{{{x^3} - {x^2}}}}}{x}} \right)^2} = {\left( {\sqrt[3]{{\frac{{{x^3} - {x^2}}}{{{x^3}}}}}} \right)^2} = {\left( {\sqrt[3]{{1 - \frac{1}{x}}}} \right)^2}$
\end{itemize}
Die nun in die Ausgangsfunktion eimsetzen:
$\mathop {\lim }\limits_{x \to \infty} \frac{x^2}{x^2\left(1+\sqrt[3]{1+\frac{1}{x}}+\left(\sqrt[3]{1+\frac{1}{x}}\right)^2\right)}=\mathop {\lim }\limits_{x \to \infty} \frac{1}{1+\sqrt[3]{1-\frac{1}{x}}+\left(\sqrt[3]{1-\frac{1}{x}}\right)^2}$
Da sich im Zähler nun kein $x$ mehr befindet, wird jetzt im Nenner $\infty$ eingesetzt. Alle $\frac{1}{x}$ werden $0$. Somit $\frac{1}{1+1+1}=\frac{1}{3}$
\end{enumerate}
%\includegraphics[scale=0.45]{gnuplot08a.png}
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xtick={-4,-3,...,4},
ytick={-5,-4,...,4},]
\addplot [mark=none,domain=-4.8:-1.05, color=NewBlue, line width=0.5mm,step=10000, smooth, tension=0.2] {1/(x^2-1)};
\addplot [mark=none,domain=-0.95:0.95, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
\addplot [mark=none,domain=1.05:4.8, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
\end{axis}
\draw[-latex,red, line width=0.75mm](3.9,2.5) node[above, left, yshift=0.5mm] {von links} -- (5.9,2.5) ;
%\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
\end{tikzpicture}
\end{tikzpicture}
%!tikz source end
\end{document}

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%!TEX root=main.tex
%\documentclass{tufte-book}
\documentclass[fleqn, a4paper,12pt]{scrbook}
%\hypersetup{colorlinks}% uncomment this line if you prefer colored hyperlinks (e.g., for onscreen viewing)
%%
% Book metadata
\title{Formeln und Notizen}
%\author[]{}
%\publisher{Ich}
%%
% If they're installed, use Bergamo and Chantilly from www.fontsite.com.
% They're clones of Bembo and Gill Sans, respectively.
%\IfFileExists{bergamo.sty}{\usepackage[osf]{bergamo}}{}% Bembo
%\IfFileExists{chantill.sty}{\usepackage{chantill}}{}% Gill Sans
\input{definitions.tex}
%%% Local Variables:
%%% TeX-master: "master"
%%% End:
\begin{document}
\tikzstyle{every picture}+=[remember picture]
% % Front matter
% \frontmatter
%
% % r.1 blank page
% \blankpage
%
% % v.2 epigraphs
%
%
% % r.3 full title page
% \maketitle
%
%
% % v.4 copyright page
% \newpage
% \begin{fullwidth}
% ~\vfill
% \thispagestyle{empty}
% \setlength{\parindent}{0pt}
% \setlength{\parskip}{\baselineskip}
% Copyright \copyright\ \the\year\ \thanklessauthor
%
% \par\smallcaps{Published by \thanklesspublisher}
%
% \par\smallcaps{tufte-latex.googlecode.com}
%
% \par Licensed under the Apache License, Version 2.0 (the ``License''); you may not
% use this file except in compliance with the License. You may obtain a copy
% of the License at \url{http://www.apache.org/licenses/LICENSE-2.0}. Unless
% required by applicable law or agreed to in writing, software distributed
% under the License is distributed on an \smallcaps{``AS IS'' BASIS, WITHOUT
% WARRANTIES OR CONDITIONS OF ANY KIND}, either express or implied. See the
% License for the specific language governing permissions and limitations
% under the License.\index{license}
%
% \par\textit{First printing, \monthyear}
% \end{fullwidth}
% r.5 contents
%\tableofcontents
%\listoffigures
% \listoftables
% r.7 dedication
\cleardoublepage
%~\vfill
%\begin{doublespace}
%\noindent\fontsize{18}{22}\selectfont\itshape
%\nohyphenation
%Dedicated to those who appreciate \LaTeX{}
%and the work of \mbox{Edward R.~Tufte}
%and \mbox{Donald E.~Knuth}.
%\end{doublespace}
%\vfill
%\vfill
% r.9 introduction
\cleardoublepage
%\chapter*{Introduction}
%
%
%
%This sample book discusses the design of Edward Tufte's
%books\cite{Tufte2001,Tufte1990,Tufte1997,Tufte2006}
%and the use of the \doccls{tufte-book} and \doccls{tufte-handout} document classes.
%
%
%%%
%% Start the main matter (normal chapters)
\mainmatter
% \begin{asydef}
% // Global Asymptote definitions can be put here.
% import three;
% usepackage("bm");
%texpreamble("\def\V#1{\bm{#1}}");
% // One can globally override the default toolbar Settings here:
% // settings.toolbar=true;
% \end{asydef}
\chapter{Übersicht über Funktionen}
% \begin{figure}
% \begin{framebox}
%
% \begin{asy}
% label("Hallo Welt");
%
% unitsize(3cm);
% size(4cm,4cm);
% import graph;
%
% real f(real x) {
% return sqrt(2*x - x^3);
% }
% draw((-4,0) -- (3,0), arrow=Arrow(HookHead));
% draw((0,-.1) -- (0,2), arrow=Arrow(HookHead));
% path g = graph(f, -3, 1.4142);
% draw(g);
% \end{asy}
% \end{framebox}
% \end{figure}
% \input{trigo01.tex}
\input{folgen01.tex}
\input{limit01.tex}
\input{uebungenlimit01.tex}
%\input{differentialrechnung01.tex}
\input{pearson_Funktionen_und_ihre_Graphen.tex}
% \input{integral01.tex}
%\input{calculus_one.tex}
% \input{formeln.tex}
%\input{FHTW.tex}
%\section{Umrechnung rad - Degree und umgekehrt}
%\begin{marginfigure}
%\input{trigon01.tikz.tex}
%\end{marginfigure}
\backmatter
% \bibliography{sample-handout}
% \bibliographystyle{plainnat}
\bibliographystyle{plain}
\bibliography{formelbib}
\printindex
\end{document}

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\begin{tikzpicture}[scale=0.75]
\pgfplotsset{compat=1.11}
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
single arrow head extend=.3cm,}}
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
%}
%\node [
% fill=blue!50, draw,
% single arrow, single arrow head indent=0ex,
% rotate=0,
% font=\sffamily
%] at (1,1.5)
%{\rotatebox{0}{ \qquad}};
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
\begin{axis}[
x=1cm,y=1cm,
axis lines=middle,
ymajorgrids=true,
xmajorgrids=true,
xmin=-5,
xmax=5,
ymin=-5,
ymax=5,
xtick={-5,-4,...,5},
ytick={-5,-4,...,5},]
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
\draw [line width=1pt,color=orange,domain=-5:5] plot(\x,{(-0--1*\x)/1});
\draw [line width=1pt,color=orange,domain=-5:5] plot(\x,{(-0-1*\x)/1});
\end{axis}
\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
\end{tikzpicture}

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\begin{tikzpicture}[scale=2]
\definecolor{mycolor1}{rgb}{0.15, 0.47, 0.70}
%\draw[->] (-1.5,-1.5) -- (2,-1.5) node[right] {$x$};
%\draw[->] (-1.5,-2.5) -- (-1.5,0.8) node[left]{\footnotesize{$y=f(x)$}};
%\clip[draw](-1.5,-1) rectangle (1.3,1);
%\clip(-1.5,-1.3) rectangle (1.5,1);
%\begin{axis} [axis lines=center]
\begin{scope}
\draw[scale=0.5,domain=-1.9:1.9,smooth,thick,variable=\x,color=mycolor1] plot ({\x},{0.75*\x^3-2*\x-1.});
%\plot [domain=-2:2, smooth, thick, color=blue] { x^3 - 2*x };
\end{scope}
%\end{axis}
\draw[-latex] (-1.5,-1.5) -- (1.8,-1.5) node[right] {\footnotesize{$x$}};
\draw[-latex] (-1.5,-1.5) -- (-1.5,0.8) node[left] {\footnotesize{$y=f(x)$}};
\draw[color=red] (-2.1,0.17) -- (1.3,0.17);
\draw[color=red] (-2.1,-1.15) -- (1.3,-1.15);
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%!TEX root=main.tex
\newpage
\section{Übungsaufgaben Grenzwerte}
\subsection{Bestimme, wie sich die Funktion $f$ im Unendlichen verhält}
\subsubsection{Aufgabe 1}
\begin{description}
\item[Verhalten gegen $+{\infty}$ ] \marginpar{Es wird nur die höchste Potenz betrachtet}$\mathop {\lim }\limits_{x \to +\infty }x^4-x^3=\mathop {\lim }\limits_{x \to +\infty }x^4=\left(+\infty\right)^4={\infty}$
\item[Verhalten gegen $-{\infty}$] $\mathop {\lim }\limits_{x \to -\infty }x^4-x^3=\mathop {\lim }\limits_{x \to -\infty }x^4=\left(-\infty\right)^4={\infty}$
\end{description}
\subsubsection{Aufgabe 2}
%http://mathenexus.zum.de/html/analysis/grenzwerte/Grenzwertplusminusunend_Ueb.htm
\begin{description}
\item[Verhalten gegen $+{\infty}$ ] $\rightarrow $ für $x>0$ \newline\newline
$f ( x ) = \frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 } = \mathop {\lim }\limits_{x \to \infty}\frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 }=\mathop {\lim }\limits_{x \to \infty}$
\item[Verhalten gegen $-{\infty}$ ]
$f ( x ) = \frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 }$
\end{description}
\paragraph{Grenzwerte in der Unendlichkeit mit Quadratwurzeln}
Da sich obige Aufgabe im negativ Unendlichen nicht so recht erklärt hat hier noch ein paar Ausführungen dazu:
\begin{itemize}
\item Wenn $ x $ positiv ist $ x=\sqrt{x^2} $ zum Beispiel, wenn $ x=3 $, dann $ x=3=\sqrt{9} $
\item Wenn $ x $ negativ ist $ x=-\sqrt{x^2} $ dem gegenüber ist also wenn $ x=-3 $ dann $ x=-3=-\sqrt{9} $
\end{itemize}
\textbf{Wichtig}: Man muss sich merken, das wenn $x=-\sqrt{x^2}$ wenn $x \rightarrow -\infty$ ist, muss man automatisch auf die negativen Werte von $x$ schauen muss.
\textbf{Beispiel 1}: $\mathop {\lim }\limits_{x \to \infty} \frac { \sqrt { 5 x ^ { 2 } + 2 x } } { x }$
Da man hier $x \rightarrow \infty$ untersucht, sind nur positive Werte von $x$ interessant und man nutzt $x=\sqrt{x^2}$.
$\lim _ { x \rightarrow \infty } \frac { \sqrt { 5 x ^ { 2 } + 2 } x } { x } = \lim _ { x \rightarrow \infty } \frac { \frac { \sqrt { 5 x ^ { 2 } + 2 } x } { \sqrt { x ^ { 2 } } } } { \frac { x } { x } } = \lim _ { x \rightarrow \infty } \frac { \sqrt { \frac { 5 x ^ { 2 } + 2 } { x ^ { 2 } } } } { 1 } = \lim _ { x \rightarrow \infty } \sqrt { 5 + \frac { 2 } { x } } = \lim _ { x \rightarrow \infty } \sqrt { 5 + 0 }=5$
\textbf{Beispiel 2}: $\mathop {\lim }\limits_{x \to \infty} \frac { \sqrt { 5 x ^ { 2 } + 2 x } } { x }$