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\pdfinfo{
   /Author (Peter Merkert, Martin Thoma)
   /Title  (Wichtige Formeln der Analysis I)
   /CreationDate (D:20120221095400)
   /Subject (Analysis I)
   /Keywords (Analysis I; Formeln)
}

\everymath={\displaystyle}

\begin{document}

\title{Analysis Formelsammlung}
\author{Peter Merkert, Martin Thoma}
\date{21. Februar 2012}

\section{Grenzwerte}
\begin{table}[ht]
\begin{minipage}[b]{0.5\linewidth}\centering

\begin{align*}
\lim_{x \to 0} \frac {\sin x}{x}  &= 1 \\
\lim_{x \to 0} \frac {e^x - 1}{x} &= 1 \\
\lim_{h \to 0} \frac {e^{{x_0} + h} - e^{x_0}}{h} &= e^{x_0} \\
\sum_{n = 0}^{\infty} (-1)^n \frac {(-1)^{n + 1}}{n} &= \log 2 \\
\cos x    &= \sum_{n = 0}^{\infty} (-1)^n \frac {x^{2n}}{(2n)!}  \\
\sin x    &= \sum_{n = 0}^{\infty} (-1)^n \frac {x^{2n + 1}}{(2n + 1)!}
\end{align*}

\end{minipage}
\hspace{0.5cm}
\begin{minipage}[b]{0.5\linewidth}
\centering

\begin{align*}
\cosh x = \frac {1}{2} (e^x + e^{-x}) &= \scriptstyle \sum_{n = 0}^{\infty} \frac {x^{2n}}{(2n)!} \\
\sinh x = \frac {1}{2} (e^x - e^{-x}) &= \sum_{n = 0}^{\infty} \frac {x^{2n + 1}}{(2n + 1)!} \\
e^x &= \sum_{n = 0}^{\infty} \frac {x^n}{n!} \\
\sum_{n = 0}^{\infty} (-1)^n \frac {x^{n + 1}}{n + 1} &= \log (1+x)   (x \in (-1,1)) \\
\sum_{n = 0}^{\infty} x^n &= \frac {1}{1 - x}    (x \in (-1,1)) \\
0,\bar{3} &= \sum_{n = 1}^{\infty} \frac {3}{(10)^n}
\end{align*}

\end{minipage}
\end{table}



\section{Zusammenhänge}
\begin{align*}
	(\cos x)^2 + (\sin x)^2 &= 1 \\
	(\cosh x)^2 - (\sinh x)^2 &= 1 \\
	\tan x  &= \frac {\sin x}{\cos x} \\
	\tanh x &= \frac {\sinh x}{\cosh x} \\
  (x + y)^n &= \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\end{align*}

\section{Ableitungen}
\begin{align*}
	(\arctan x)' &= \frac {1}{1 + x^2} \\
	(\sin x)'    &= \cos x \\
	(\cos x)'    &= -\sin x \\
	(\arctanh x)' &= \frac {1}{\sqrt {1 + x^2}}
\end{align*}


\section{Potenzreihen}
Zuerst den Potenzradius r berechnen:
\(
	r = \frac {1}{\lim \text{sup} \sqrt[n]{|a_n|}}
\)

\end{document}