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\begin{document}
	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 2 \medskip\\}
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	{
		\renewcommand{\theenumi}{(\alph{enumi})}
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		\begin{prob*} Solve the following limits:
			\begin{enumerate}
				\item $\lim\limits_{h\to 0}\left(\dfrac{\sqrt{1+h}-1}{h}\right)$
				\item $\lim\limits_{x\to 2}\left(\dfrac{x^3+x^2-4x-4}{x-2}\right)$
			\end{enumerate}
		\end{prob*}
		\begin{soln*}
			\mbox{}
			\begin{enumerate}
				\item \begin{align*}
						\lim\limits_{h\to 0}\left(\dfrac{\sqrt{1+h}-1}{h}\right) &= 	\lim\limits_{h\to 0}\Bigg(\left(\dfrac{\sqrt{1+h}-1}{h}\right)\left(\dfrac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\right)\Bigg) \\
					&=	\lim\limits_{h\to 0}\left(\dfrac{1+h-1}{h(\sqrt{1+h}+1)}\right)\\
					&=	\lim\limits_{h\to 0}\left(\dfrac{h}{h(\sqrt{1+h}+1)}\right)\\
					&=	\lim\limits_{h\to 0}\left(\dfrac{1}{\sqrt{1+h}+1}\right)\\
					&=	\dfrac{1}{\sqrt{\vphantom{I}1+0}+1}\\
					&=	\dfrac{1}{2}
					\end{align*}
				\item We divide.
				\begin{equation*}
				\polylongdiv{x^3+x^2-4x-4}{x-2}
				\end{equation*}
				Then we can evaluate the limit fairly easily.
				\begin{align*}
					\lim\limits_{x\to 2}\left(\dfrac{x^3+x^2-4x-4}{x-2}\right) & = \lim\limits_{x\to 2}\left(\dfrac{(x-2)(x^2+3x+2)}{x-2}\right) \\
					& = \lim\limits_{x\to 2}\left(x^2+3x+2\right) \\
					& = 12
				\end{align*}
				Alternatively, instead of division, in this problem it is actually factorable.
				\begin{align*}
					x^3+x^2-4x-4 &= x^2(x+1) - 4(x+1) \\
					&= (x+1)(x^2 - 4) \\
					&= (x+2)(x-2)(x+1)
				\end{align*}



			\end{enumerate}

		\end{soln*}

\end{document}