\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}