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

\subsubsection*{Solution}

This is not in one of the forms \( 0/0 \) or \( \infty /\infty  \); it has
to be transformed into one of those forms. For this one, the trick is to conjugate;
that is, multiply numerator and denominator by the sum of those square roots.
This trick was used before, in an example in the section on finding the derivative
by the definition. ../Unit2/derivative-definition-ex4-sol.html \begin{eqnarray*}
\lim _{x\rightarrow \infty }\left( \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x}\right)  & = & \lim _{x\rightarrow \infty }\frac{\left( \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x}\right) \left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) }{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) }\\
 & = & \lim _{x\rightarrow \infty }\frac{(x^{2}+x+1)-(x^{2}-x)}{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) }\\
 & = & \lim _{x\rightarrow \infty }\frac{2x+1}{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) }.
\end{eqnarray*}
However, once we have done that, we no longer need l'H\"{o}pital's rule! All
we need to do here is divide out top and bottom by the largest common power
of \( x \) (in this case, \( x \) itself), and bring that power inside the
square roots in the denominator:\begin{eqnarray*}
\lim _{x\rightarrow \infty }\left( \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x}\right)  & = & \lim _{x\rightarrow \infty }\frac{\left( 2x+1\right) \frac{1}{x}}{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) \frac{1}{x}}\\
 & = & \lim _{x\rightarrow \infty }\frac{2+\frac{1}{x}}{\left( \sqrt{1+\frac{1}{x}+\frac{1}{x^{2}}}+\sqrt{1-\frac{1}{x}}\right) }.
\end{eqnarray*}
It is now standard to see that the two terms in the denominator each goes to
1, and the numerator goes to 2, leaving a final limit of 1,\[
\lim _{x\rightarrow \infty }\left( \sqrt{x^{2}+x+1}-\sqrt{x^{2}-x}\right) =1.\]
Now, you could apply l'H\"{o}pital's rule to \[
\lim _{x\rightarrow \infty }\frac{2x+1}{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) },\]
 but it just doesn't help:\begin{eqnarray*}
\lim _{x\rightarrow \infty }\frac{2x+1}{\left( \sqrt{x^{2}+x+1}+\sqrt{x^{2}-x}\right) } & = & \lim _{x\rightarrow \infty }\frac{2}{\left( \frac{2x+1}{2\sqrt{x^{2}+x+1}}+\frac{2x-1}{2\sqrt{x^{2}-x}}\right) },
\end{eqnarray*}
leaving you with two limits that are essentially as complicated as where we
were before. This often happens with l'H\"{o}pital's rule problems, not every
problem is actually simplified by this technique.
\end{document}
