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authorKarl Berry <karl@freefriends.org>2010-07-09 00:20:12 +0000
committerKarl Berry <karl@freefriends.org>2010-07-09 00:20:12 +0000
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-\documentclass{article}
-\usepackage{amsmath}
-\usepackage[myconfig,forpaper,pointsonleft,nosolutions]{eqexam}
-
-\examNum{1}
-\forVersion a
-\VersionAtext{Quiz~\nExam--003}
-\VersionBtext{Quiz~\nExam--007}
-\shortVersionAtext{Q{\nExam}s3}
-\shortVersionBtext{Q{\nExam}s7}
-
-\title[\sExam]{\bfseries\Exam}
-\author{D. P. Story}
-\subject[C1]{Calculus I}
-\date{Spring \the\year}
-\keywords{Test~\nExam, Section \ifAB{003}{007}}
-\email{dpstory@uakron.edu}
-
-\everymath{\displaystyle}
-
-\begin{document}
-
-\maketitle
-
-\begin{exam}{Part1}
-
-\begin{instructions}[Instructions:]
-Solve each of the following problems without error. \textit{Show all details.} Box in your
-$\boxed{\text{answers}}$. Use good notation, you \emph{will} be marked off for bad notation.
-\textbf{Note:} The value of a limit can be a number, the symbol $+\infty$, the symbol $-\infty$,
-or may be labelled DNE (for ``does not exist'').
-\end{instructions}
-
-\begin{problem}[4]
-Compute $ \ifAB{\lim_{x\to-1}\frac{4x^2+x}{x}}{\lim_{x\to2}\frac{1-3x}{x+1}}$
-\begin{solution}[2in]
-As discussed in class, this is a ``Skill Level 0'' limit problem:
-$$
-\ifAB{\lim_{x\to-1}\frac{4x^2+x}{x}}{\lim_{x\to2}\frac{1-3x}{x+1}}
- = \ifAB{\frac{4(-1)^2+(-1)}{-1}}{\lim_{x\to2}\frac{1-3(2)}{2+1}}
- = \boxed{\ifAB{-3}{-\frac{5}{3}}}
-$$
-\end{solution}
-\end{problem}
-
-\begin{problem}[3]
-Define the function $ f(x) = \begin{cases} 2x^3 - 1 & x < -2\\ 2- x^2 & x \ge -2\end{cases}$.
-Compute $\lim_{x\to\ifAB{-2^-}{-2^+}} f(x) $, show the details of your reasoning.
-
-\begin{solution}[2in]
-We use standard techniques:
-\begin{verA}
-\begin{alignat*}{2}
- \lim_{x\to-2^-} f(x) &
- = \lim_{x\to-2^-} (2x^3-1) &&\qquad\text{since $ x < -2$}\\&
- = 2(-2)^3 - 1&&\qquad\text{now a skill level 0 problem}\\&
- = \boxed{-17}
-\end{alignat*}
-\end{verA}
-\begin{verB}
-\begin{alignat*}{2}
- \lim_{x\to-2^+} f(x) &
- = \lim_{x\to-2^+} (2- x^2) &&\qquad\text{since $ x < -2$}\\&
- = 2 - (-2)^2&&\qquad\text{now a skill level 0 problem}\\&
- = \boxed{-2}
-\end{alignat*}
-\end{verB}
-\end{solution}
-\end{problem}
-
-\begin{problem}[3]
-Compute $\ifAB{\lim_{x\to2} \frac{1-x}{(x-2)^2}}
- {\lim_{x\to3} \frac{x-2}{(3-x)^2}}$
-
-\begin{solution}[1in]
-\begin{verA}
-Notice the denominator goes to zero, but the numerator does not;
-this indicates a vertical asymptote usually. Because the
-denominator is squared, it's always positive. When $x$ is
-``close'' to $2$, $1 - x < 0$, that is, when $x$ is ``close'' to
-$2$ the numerator is \emph{negative}. The ratio of the numerator and
-denominator is \emph{negative} when $x$ is ``close'' to $2$. Thus, we
-conclude,
-$$
- \boxed{\lim_{x\to2} \frac{1-x}{(x-2)^2} = -\infty}
-$$
-\end{verA}
-\begin{verB}
-Notice the denominator goes to zero, but the numerator does not;
-this indicates a vertical asymptote usually. Because the
-denominator is squared, it's always positive. When $x$ is
-``close'' to $3$, $x - 2 > 0$, that is, when $x$ is ``close'' to
-$3$ the numerator is \emph{positive}. The ratio of the numerator and
-denominator is \emph{positive} when $x$ is ``close'' to $3$. Thus, we
-conclude,
-$$
- \boxed{\lim_{x\to3} \frac{x-2}{(3-x)^2} = +\infty}
-$$
-\end{verB}
-\end{solution}
-\end{problem}
-
-\end{exam}
-\end{document}