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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{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{proof}
This result uses the Generalized MVT from earlier in this chapter. Let's deal
with limits on the right. Assume that \[
\lim _{x\rightarrow a+}f(x)=0,\, \, \mathrm{and}\, \, \lim _{x\rightarrow a+}g(x)=0,\]
and that \( f \) and \( g \) are differentiable on some interval \( (a,b). \)
If you set \( g(a)=f(a)=0 \), then the functions \( f \) and \( g \) will
be continuous on \( [a,b] \) and differentiable on \( (a,b) \), so will fit
the hypotheses of the generalized MVT. Now, there should be some question about
the hypothesis that \( g'(x)\neq 0 \), but certainly the limit \[
\lim _{x\rightarrow a+}\frac{f'(x)}{g'(x)}\]
won't make sense unless \( g'(x)\neq 0 \) on some interval near \( a \), so
we can choose \( b \) so that \( g'(x)\neq 0 \) on \( (a,b) \). Then, by
the generalized MVT, if \( x\in (a,b) \), \begin{eqnarray*}
\frac{f(x)-f(a)}{g(x)-g(a)} & = & \frac{f(x)}{g(x)}\\
 & = & \frac{f'(c)}{g'(c)},
\end{eqnarray*}
for some \( c \) between \( a \) and \( x \). Since \( c \) is trapped between
\( a \) and \( x \), as \( x \) approaches \( a \), \( c \) will also approach
\( a \), and so \begin{eqnarray*}
\lim _{x\rightarrow a+}\frac{f(x)}{g(x)} & = & \lim _{c\rightarrow a+}\frac{f'(c)}{g'(c)}\\
 & = & \lim _{x\rightarrow a+}\frac{f'(x)}{g'(x)}.
\end{eqnarray*}
Since the same idesa work out for the limits from the left, the statement follows.\end{proof}

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