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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}

This is a nice, computational formula that can be used to compute limits, and
so is often taught in high-school calculus classes. It usually isn't justified
there, however. It also isn't usually mentioned how the name came about. In
the \( 17^{th} \) and \( 18^{th} \) centuries, it was common for kings and
other rulers, rather than universities, to employ mathematicians. l'H\^{o}pital
http://www.britannica.com/bcom/eb/article/7/0,5716,80997+1+78865,00.html or
http://www-gap.dcs.st-and.ac.uk/\textasciitilde{}history/Mathematicians/De\_L'Hopital.html
was one such employer, who hired a mathematician by the name of Johann Bernoulli
to prove him a theorem. Bernoulli did, and the result was named for the nobleman
who commissioned the work rather than the mathematician who thought it up. 

l'H\^{o}pital's Rule is used to make sense out of limits that are \textit{indeterminate},
that is, limits that you can't figure out by just plugging in the limiting value
of \( x \) into the expression. We have seen many such limits, usually of the
form \[
\frac{0}{0},\]
 that is, a limit: \[
\lim _{x\rightarrow a}\frac{f(x)}{g(x)}\]


where both \[
\lim _{x\rightarrow a}f(x)=0\]
 and \[
\lim _{x\rightarrow a}g(x)=0.\]
 \( a \) here could be a number, or \( \pm \infty  \).  It's important to
make the distinction between making sense of a \emph{limit} of the form \( 0/0 \)
versus actually making sense of the \emph{fraction} itself. \( 0/0 \) is \emph{only}
understandable as a limit.

A limit is \textit{indeterminate} if it is one of the following forms (that
is, the pieces of the expression have the indicated limits themselves): \[
\frac{0}{0},\, \, \frac{\infty }{\infty },\, \, 1^{\infty },\, \, \infty -\infty ,\, \, 0\times \infty ,\, \, 0^{0},\, \, 0^{\infty },\, \, \mathrm{or}\, \, \infty ^{0.}\]


By some standard algebraic tricks, any of the other forms can be reduced to
one of the first two (though the last two should be kept separate).  

Here is what l'H\^{o}pital paid for: 

\begin{thm}
\textbf{\textup{{[}l'H\^{o}pital's Rule, 0/0 form{]}.}} Let \( f \) and \( g \)
be differentiable on some interval, with \( a \) in that interval, and assume
they satisfy \[
\lim _{x\rightarrow a}f(x)=0,\, \, \mathrm{and}\, \, \lim _{x\rightarrow a}g(x)=0.\]
 Then, \[
\lim _{x\rightarrow a}\frac{f(x)}{g(x)}=\lim _{x\rightarrow a}\frac{f'(x)}{g'(x)}.\]

\end{thm}
\begin{rem}
\( a \) could be a number, or it could be \( \infty  \). The theorem also
holds for one-sided limits, like \[
\lim _{x\rightarrow a+}\frac{f(x)}{g(x)},\]
 as well.
\end{rem}
\textbf{\emph{Proof.}}

\begin{example}
\[
\lim _{x\rightarrow 0}\frac{\sin (3x)}{\tan (4x)}\]
 
\end{example}

\subsubsection*{Solution}

\begin{xca}
\[
\lim _{x\rightarrow 2}\frac{\sqrt{x}-2}{x^{2}-3x+2}=\]

\end{xca}
~

\begin{xca}
\[
\lim _{x\rightarrow 0}\frac{\cos (x)-1}{x^{2}}=\]
 
\end{xca}

\subsubsection*{Hint:}

Here, once you apply l'H\"{o}pital's rule, you will still be left with an indeterminate
limit. Apply l'H\"{o}pital's rule \emph{again}. You can continue to do this
until you come to a fraction you can simplify, or until the limit is no longer
an indeterminate form.


\subsubsection{Infinity/infinity}

There is a separate l'H\"{o}pital's rule theorem for for \( \infty /\infty  \)
--type limits, \[
\lim _{x\rightarrow a}\frac{f(x)}{g(x)},\, \textrm{where}\, \lim _{x\rightarrow a}f(x)=\infty =\lim _{x\rightarrow a}g(x).\]
 But the statement is essentially the same. The proof is a bit more subtle;
it can't \emph{quite} be solved by turning it into the limit \[
\lim _{x\rightarrow a}\frac{1/g(x)}{1/f(x)},\]
since that trick requires that you know that the limit exists in order to show
how to compute it. l'H\"{o}pital's rule actually does not require in advance
that you know the limit exists. It says that the limit exists and equals the
limit of the quotient of the derivatives, if that latter limit exists. 

\begin{thm*}
\textbf{\textup{{[}l'H\^{o}pital's Rule, \( \infty /\infty  \) form{]}.}} Let
\( f \) and \( g \) be differentiable on some interval, with \( a \) in that
interval, and assume they satisfy \[
\lim _{x\rightarrow a}f(x)=\infty ,\, \, \mathrm{and}\, \, \lim _{x\rightarrow a}g(x)=\infty .\]
 Then, \[
\lim _{x\rightarrow a}\frac{f(x)}{g(x)}=\lim _{x\rightarrow a}\frac{f'(x)}{g'(x)}.\]

\end{thm*}
\begin{example}
\[
\lim _{x\rightarrow \infty }\frac{x^{3}}{e^{x}}\]
 
\end{example}

\subsubsection*{Solution}

\begin{example}
\[
\lim _{x\rightarrow \infty }\frac{\ln (x)}{x^{2}}\]

\end{example}

\subsubsection*{Solution}



\begin{example}
\[
\lim _{x\rightarrow \infty }\left( \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x}\right) \]
 
\end{example}

\subsubsection*{Solution}

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

\end{example}

\subsubsection*{Solution}

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

\end{xca}

\subsubsection{Growth rates of functions.}

For functions that go to infinity as \( x\rightarrow \infty  \), we can use
l'H\"{o}pital's rule to compare how quickly they go to infinity. We say that
\( f<<g \) (or, \( g>>f \)) if \[
\lim _{x\rightarrow \infty }\frac{f(x)}{g(x)}=0,\]
 or, equivalently, \[
\lim _{x\rightarrow \infty }\frac{g(x)}{f(x)}=\infty .\]


As an example, since \begin{eqnarray*}
\lim _{x\rightarrow \infty }\frac{e^{x}}{x^{2}} & = & \lim _{x\rightarrow \infty }\frac{e^{x}}{2x}\\
 & = & \lim _{x\rightarrow \infty }\frac{e^{x}}{2}\\
 & = & \infty ,
\end{eqnarray*}
we say that \( e^{x}>>x^{2} \), as you might expect.

\begin{xca}
Place the following functions in order from the slowest growth rate to the fastest,
as \( x\rightarrow \infty  \). \[
e^{x},\, x^{2},\, \ln (x),\, x^{x},\, x\ln (x),\, e^{x}\ln (x).\]
You do not have to use l'H\"{o}pital's rule in every case. 

\textbf{\emph{Answer:}}\end{xca}

\end{document}
