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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{l'H\^{o}pital's Rule}

\begin{example}
\[
\lim _{x\rightarrow \infty }\left( 1+\frac{1}{x}\right) ^{x}\]

\end{example}

\subsubsection*{Solution}

Again, this is not in one of the forms \( 0/0 \) or \( \infty /\infty  \);
it has to be transformed into one of those forms. 

First we deal with the natural logarithm of this expression. Then we will use
the fact that \[
\lim _{x\rightarrow a}e^{f(x)}=e^{\lim _{x\rightarrow a}f(x)},\]
which is true because the exponential is a continuous function href=\char`\"{}../Unit1/continuity-theory.html\char`\"{},
to show the limit we are really after.\begin{eqnarray*}
\lim _{x\rightarrow \infty }\ln \left( \left( 1+\frac{1}{x}\right) ^{x}\right)  & = & \lim _{x\rightarrow \infty }x\ln \left( 1+\frac{1}{x}\right) \\
 & = & \lim _{x\rightarrow \infty }\frac{\ln \left( 1+\frac{1}{x}\right) }{\frac{1}{x}}.
\end{eqnarray*}
The first step there just brought the exponent out of the logarithm, using the
standard rules for logarithms href=\char`\"{}Unit3/exponential-basics.html\char`\"{}.
This changed the type of indeterminate form from \( 1^{\infty } \) to \( \infty \cdot 0 \).
Then, in a more standard step, the \( \infty \cdot 0 \) form is converted to
\( 0/0 \). Now, at that stage there was a choice, to put one or the other of
those terms in the denominator. But it makes more sense to put the simpler term
in the denominator, since you have to invert it. Inverting a complicated expression
gives you a mess to differentiate. Inverting a simple expression is not as likely
to cause trouble.

Continuing, applying l'H\"{o}pital's rule and not forgetting the chain rule,
\begin{eqnarray*}
\lim _{x\rightarrow \infty }\ln \left( \left( 1+\frac{1}{x}\right) ^{x}\right)  & = & \lim _{x\rightarrow \infty }\frac{\ln \left( 1+\frac{1}{x}\right) }{\frac{1}{x}}\\
 & = & \lim _{x\rightarrow \infty }\frac{\left( \frac{1}{1+\frac{1}{x}}\right) \left( 0-\frac{1}{x^{2}}\right) }{-\frac{1}{x^{2}}}\\
 & = & \lim _{x\rightarrow \infty }\frac{\left( \frac{1}{1+\frac{1}{x}}\right) \left( 1\right) }{1}\\
 & = & 1.
\end{eqnarray*}
But we're not yet done. We found the limit of the logarithm. To get back to
the original limit, \begin{eqnarray*}
\lim _{x\rightarrow \infty }\left( 1+\frac{1}{x}\right) ^{x} & = & \lim _{x\rightarrow \infty }e^{\left( \ln \left( 1+\frac{1}{x}\right) ^{x}\right) }\\
 & = & e^{\left( \lim _{x\rightarrow \infty }\ln \left( \left( 1+\frac{1}{x}\right) ^{x}\right) \right) }\\
 & = & e^{\left( 1\right) }\\
 & = & e.
\end{eqnarray*}

\end{document}
