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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 7\medskip\\}
	\hrule
	{
		\begin{prob*}
			\mbox{}
			\begin{enumerate}
				\item $\lim\limits_{x\to\infty} x^{\frac{\ln(2)}{1+\ln(x)}}$
				\item $\lim\limits_{x\to 0} \csc(x)-\cot(x)$
			\end{enumerate}
		\end{prob*}
		\begin{soln*}
			\mbox{}
			\begin{enumerate}
				\item We begin by taking the logarithm, as this turns into an indeterminate form (namely $\infty^0$).
				\begin{align*}
					\lim\limits_{x\to\infty} \ln(x^{\frac{\ln(2)}{1+\ln(x)}}) &= \lim\limits_{x\to\infty}\frac{\ln(2)}{1+\ln(x)}\cdot \ln(x) \\
					&= \ln(2) \cdot \lim\limits_{x\to\infty}\frac{\ln(x)}{1+\ln(x)}
				\end{align*}
				We now can use L'H\^opital's Rule.
				\begin{align*}
					\ln(2) \cdot \lim\limits_{x\to\infty}\frac{\ln(x)}{1+\ln(x)} &= \ln(2) \cdot \lim\limits_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{x}}\\ &= \ln(2)
				\end{align*}
				Remember, you took a log. Therefore, the solution is $e^{\ln(2)}$, which is simply $2$
				\item
				\begin{align*}
					\lim\limits_{x\to 0} \csc(x)-\cot(x) &= \lim\limits_{x\to 0} \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} \\
					&= \lim\limits_{x\to 0} \frac{1-\cos(x)}{\sin(x)}\\
					&= \lim\limits_{x\to 0} \frac{\sin(x)}{\cos(x)}\\
					&= 0
				\end{align*}

			\end{enumerate}
		\end{soln*}

	}
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