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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{Trigonometric functions}


\subsubsection{Limits of trig functions}

\begin{example}
\[
\displaystyle \lim _{x\rightarrow 0}\frac{\cos (x)-1}{x}=0\]

\end{example}

\subsubsection*{Solution}

We could start over, going through the same sort of geometric-trigonometric
arguments that we did for the limit of \( \sin (x) \). But, instead, we can
be trickier.\begin{eqnarray*}
\frac{\cos (x)-1}{x} & = & \frac{\left( \cos (x)-1\right) \left( \cos (x)+1\right) }{x\left( \cos (x)+1\right) }\\
 & = & \frac{\cos ^{2}(x)-1}{x\left( \cos (x)+1\right) }\\
 & = & \frac{-\sin ^{2}(x)}{x(\cos (x)+1)}.
\end{eqnarray*}
Now, after that trick, the numerator has been transformed into \( \sin (x) \)
(well, two of them). But one of those \( \sin (x) \)'s can be used to cover
the \( x \) in the denominator:\begin{eqnarray*}
\lim _{x\rightarrow 0}\frac{\cos (x)-1}{x} & = & \lim _{x\rightarrow 0}\frac{-\sin ^{2}(x)}{x(\cos (x)+1)}\\
 & = & -\lim _{x\rightarrow 0}\left( \frac{\sin (x)}{x}\right) \left( \frac{\sin (x)}{(\cos (x)+1)}\right) \\
 & = & -\left( \lim _{x\rightarrow 0}\frac{-\sin ^{2}(x)}{x(\cos (x)+1)}\right) \left( \lim _{x\rightarrow 0}\frac{\sin (x)}{(\cos (x)+1)}\right) \\
 & = & -\left( 1\right) \frac{0}{2}\\
 & = & 0.
\end{eqnarray*}

\end{document}
