\begin{frame}
  \frametitle{Continuously Compounded Interest}

  \vspace{-.5ex}
  \begin{block}{}
    If the interest is compounded $n$ times per year,
    then after $t$ years the value is\vspace{-2.5ex}
    \begin{talign}
      A_0 \cdot \left(1+\frac{r}{n}\right)^{nt}
    \end{talign}
  \end{block}
  \pause
  
  \begin{exampleblock}{}
    For instance, $1000\$$ with 6\% interest after $3$ years:
    \begin{itemize}
    \pause
      \item $1000\$ \cdot (1 + 0.06)^3 = 1191.02\$$ annual compounding
    \pause
      \item $1000\$ \cdot (1 + 0.03)^6 = 1194.05\$$ semiannual compounding
    \pause
      \item $1000\$ \cdot (1 + 0.015)^{12} = 1195.62\$$ quarterly compounding
    \pause
      \item $1000\$ \cdot (1 + 0.005)^{36} = 1196.68\$$ monthly compounding
    \pause
      \item $1000\$ \cdot (1 + 0.06/356)^{356\cdot 3} = 1197.20\$$ daily compounding
    \end{itemize}
  \end{exampleblock}
  \pause
  
  \begin{block}{}
  If we let $n \to \infty$, we get \emph{continuous compounding}:\vspace{-1ex}
  \begin{talign}
    A(t) = \lim_{n\to \infty} A_0 \cdot \left(1+\frac{r}{n}\right)^{nt}
    \mpause[1]{= A_0 \cdot \left( \lim_{n\to \infty} \left(1+\frac{r}{n}\right)^\frac{n}{r} \right)^{rt} }
    \mpause[2]{= A_0 \cdot e^{rt} }
  \end{talign}
  \end{block}
    \pause\pause\pause
  \begin{exampleblock}{}
    \begin{itemize}
      \item $1000\$ \cdot e^{0.06\cdot 3} = 1197.22\$$ continuous compounding
    \end{itemize}
  \end{exampleblock}
\end{frame}