%Used: Math 99 Sample Exam 3
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\lhead{\color{blue} \Large Lecture Notes}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,    2013}
\rfoot{\small   Last revised: October 18,    2013}
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\chead{\LARGE Limits involving $e$ }
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\begin{document}


\fbox{~~Definition: \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n%
}\right) ^{n}=e~~$}\bigskip

\ \ Theorem: \ $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{a}{n}%
\right) ^{n}=e^{a}$\bigskip

\ \ Proof: \ If $a=0$, then clearly $\lim\limits_{n\rightarrow \infty
}\left( 1+\dfrac{a}{n}\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left(
1+\dfrac{0}{n}\right) ^{n}=\lim\limits_{n\rightarrow \infty }1^{n}=1$. \ If $%
a\not=0$, then%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{a}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n}{a}~~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n%
}{a}~~}\right) ^{\tfrac{n}{a}\cdot a}=\lim\limits_{n\rightarrow \infty }%
\left[ \left( 1+\dfrac{1}{~~\dfrac{n}{a}~~}\right) ^{\tfrac{n}{a}}\right]
^{a}
\end{equation*}%
\bigskip

Define $m=\dfrac{n}{a}$. \ \ Since $a$ is fixed, $m$ approaches infinity (or
negative infinity if $a$ is negative) as $n$ approaches infinity.%
\begin{equation*}
\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n}{a}~~}%
\right) ^{\tfrac{n}{a}}\right] ^{a}=\left[ \lim\limits_{m\rightarrow \infty
}\left( 1+\dfrac{1}{m}\right) ^{m}\right] ^{a}=e^{a}
\end{equation*}%
{\LARGE \bigskip \bigskip }\bigskip

\begin{center}
{\LARGE Practice Problems\bigskip }
\end{center}

Assume that%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}=e\text{
\ and \ }\lim\limits_{n\rightarrow -\infty }\left( 1+\dfrac{1}{n}\right)
^{n}=e
\end{equation*}%
and compute each of the following limits.\bigskip {\LARGE \bigskip }%
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\begin{enumerate}
\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}\medskip \medskip $

\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}\medskip \medskip $

\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}\medskip \medskip $

\item $\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) ^{n}$ $%
\medskip \medskip $

\item $\lim\limits_{n\rightarrow \infty }\left( \dfrac{n}{n+2}\right) ^{n}$
\end{enumerate}

\bigskip

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\pagebreak

\begin{center}
{\LARGE Answers - Practice Problems\bigskip }
\end{center}

1.) $\ e\medskip \medskip $ \ \qquad 2.) $\ e^{2}\medskip \medskip \qquad \
\ $ 3.) $\ e^{5}\medskip \medskip $ \qquad\ \ 4.) $\ \dfrac{1}{e}\qquad \ \
\ $5.) $\ \dfrac{1}{e^{2}}$

\bigskip

\begin{center}
{\LARGE Solutions - Practice Problems\bigskip }
\end{center}

Assume that%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}=e\text{
\ and \ }\lim\limits_{n\rightarrow -\infty }\left( 1+\dfrac{1}{n}\right)
^{n}=e
\end{equation*}%
and compute each of the following limits.\bigskip

\begin{enumerate}
\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}\medskip \medskip $%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n+2}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{n}\left( 1+\dfrac{1}{n}\right) ^{2}=\lim\limits_{n\rightarrow \infty
}\left( 1+\dfrac{1}{n}\right) ^{n}\lim\limits_{n\rightarrow \infty }\left( 1+%
\dfrac{1}{n}\right) ^{2}=e\cdot 1=e
\end{equation*}

\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}\medskip \medskip $%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right)
^{2n}=\lim\limits_{n\rightarrow \infty }\left[ \left( 1+\dfrac{1}{n}\right)
^{n}\right] ^{2}=\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{%
n}\right) ^{n}\right] ^{2}=e^{2}
\end{equation*}

\item $\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}\medskip \medskip $%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{5}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n}{5}~~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n%
}{5}~~}\right) ^{\tfrac{n}{5}\cdot 5}=\lim\limits_{n\rightarrow \infty }%
\left[ \left( 1+\dfrac{1}{~~\dfrac{n}{5}~~}\right) ^{\tfrac{n}{5}}\right]
^{5}
\end{equation*}%
Let $m=\dfrac{n}{5}$. \ As $n$ approaches infinity, so does $m$.%
\begin{equation*}
\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n}{5}~~}%
\right) ^{\tfrac{n}{5}}\right] ^{5}=\left[ \lim\limits_{m\rightarrow \infty
}\left( 1+\dfrac{1}{m}\right) ^{m}\right] ^{5}=e^{5}~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation*}%
\pagebreak

\item $\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) ^{n}$ $%
\medskip \medskip $%
\begin{eqnarray*}
\lim\limits_{n\rightarrow \infty }\left( 1-\dfrac{1}{n}\right) ^{n}
&=&\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{-1}{n}\right)
^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n}{-1}~~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{~~\dfrac{n%
}{-1}~~}\right) ^{\tfrac{n}{-1}\cdot \left( -1\right) } \\
&=&\lim\limits_{n\rightarrow \infty }\left[ \left( 1+\dfrac{1}{-n}\right)
^{\left( -n\right) }\right] ^{\left( -1\right) }=\left[ \lim\limits_{n%
\rightarrow \infty }\left( 1+\dfrac{1}{-n}\right) ^{\left( -n\right) }\right]
^{\left( -1\right) }
\end{eqnarray*}%
\medskip

Let $m=-n$%
\begin{equation*}
\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{-n}\right)
^{\left( -n\right) }\right] ^{\left( -1\right) }=\left[ \lim\limits_{m%
\rightarrow -\infty }\left( 1+\dfrac{1}{m}\right) ^{m}\right] ^{\left(
-1\right) }=e^{-1}=\dfrac{1}{e}
\end{equation*}

\item $\lim\limits_{n\rightarrow \infty }\left( \dfrac{n}{n+2}\right) ^{n}$%
\begin{eqnarray*}
\lim\limits_{n\rightarrow \infty }\left( \dfrac{n}{n+2}\right) ^{n}
&=&\lim\limits_{n\rightarrow \infty }\left( \dfrac{1}{~~\dfrac{n+2}{n}~~}%
\right) ^{n}=\lim\limits_{n\rightarrow \infty }\left( \left( \dfrac{n+2}{n}%
\right) ^{\left( -1\right) }\right) ^{n}=\lim\limits_{n\rightarrow \infty }%
\left[ \left( 1+\dfrac{2}{n}\right) ^{n}\right] ^{\left( -1\right) } \\
&=&\left[ \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{2}{n}\right) ^{n}%
\right] ^{\left( -1\right) }=\left( e^{2}\right) ^{\left( -1\right) }=e^{-2}=%
\dfrac{1}{e^{2}}
\end{eqnarray*}
\end{enumerate}

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For more documents like this, visit our page at\
https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
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