% Time-stamp: <test.tex: Last changed on 26-01-17 16:35:40 cet by mike@moocow>
% Added packages eurosym (¤) and inputenc (not quite sure what it does; wanted ec fonts)
\documentclass[a4paper,twoside,12pt]{amsart}
\usepackage[german,english]{babel}
\usepackage{amsmath}
\usepackage{amstext}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{amscd}
\begin{document}

\date{\today}

%\tableofcontents

\begin{displaymath}
  f(x) = \left(1+{3\over\sqrt(x)}\right)^{2\sqrt x}
\end{displaymath}
Erst ersetze \(\sqrt x\) mit \(u\).  Das ist stetig und monoton f\"ur \(x>0\).

\begin{displaymath}
  f(u)=\left(1+{3\over u}\right)^{2u}.
\end{displaymath}
Dann Betrachte \(\ln(f(u))\).  Als \(u\to\infty\), \(\ln\) ist monoton und stetig!

\begin{displaymath}
  \lim_{u\to\infty}2u\ln\left(1+{3\over u}\right)=\lim_{u\to\infty}2{{\left(\ln(1+{3\over
      u}\right)}\over u^{-1}}
\end{displaymath}
L'H\^opital:
\begin{displaymath}
  \lim_{u\to\infty}{6\over1+{3\over u}}=6
\end{displaymath}
Also \(\lim_{u\to\infty}\ln(f(u))=6,\quad\lim_{x\to\infty}\ln(f(x))=36,\qquad\lim_{x\to\infty}f(x)=e^{36}\).
\end{document}

\endinput