01_03_ex_39.tex

{Theorem \ref{thm:special_limits} states that $\ds \lim_{x\to0} \frac{\cos x-1}{x}=0$. Work through the following steps to establish this result.
\begin{enumerate}
\item	Show why $\ds \frac{\cos x-1}{x} = \frac{-\sin^2 x}{x(\cos x+1)}$. Hint: multiply $\ds \frac{\cos x-1}{x}$ by 1, written in special form.
\item	Evaluate $\ds \lim_{x\to 0} \frac{-\sin^2 x}{x(\cos x+1)}$ by breaking the expression $\ds \frac{-\sin^2 x}{x(\cos x+1)}$ into a product of two functions and using Theorem \ref{thm:special_limits}.
\item		How do the above steps establish that $\ds \lim_{x\to0} \frac{\cos x-1}{x}=0$?
\end{enumerate}}
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