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\begin{frame}
  \frametitle{The Number $e$ as a Limit}
  
  Let $f(x) = \ln x$. We know that
  \begin{talign}
    f'(x) = \frac{1}{x} &&\mpause[1]{\text{ and hence }}&&\mpause[1]{f'(1) = 1}
  \end{talign}
  \pause\pause
  By definition of the limit
  \begin{talign}
    1 = f'(1) &= \lim_{h\to 0} \frac{f(1+h) - f(1)}{h}
    \mpause[1]{= \lim_{h\to 0} \frac{\ln(1+h) - \ln(1)}{h}} \\
    &\mpause[2]{= \lim_{h\to 0} \left( \frac{1}{h}\cdot \ln(1+h) \right)} 
    \mpause[3]{= \lim_{h\to 0} \ln(1+h)^{\frac{1}{h}}} 
  \end{talign}
  \pause\pause\pause\pause
  As a consequence we get
  \begin{talign}
    e 
    \mpause[1]{= e^1}
    \mpause[2]{= e^{f'(1)} }
    \mpause[3]{= e^{\lim_{h\to 0} \ln(1+h)^{\frac{1}{h}}} }
    \mpause[4]{= \lim_{h\to 0} e^{\ln(1+h)^{\frac{1}{h}}} }
    \mpause[5]{= \lim_{h\to 0} (1+h)^{\frac{1}{h}} }
  \end{talign}
  \pause\pause\pause\pause\pause\pause  
  \begin{block}{}
    \begin{malign}
      e \quad=\quad \lim_{h\to 0} (1+h)^{\frac{1}{h}}
      \mpause[1]{\quad=\quad \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n}  }
    \end{malign}
  \end{block}
\end{frame}