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author | Karl Berry <karl@freefriends.org> | 2010-07-09 00:20:12 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2010-07-09 00:20:12 +0000 |
commit | e30a75f6124a943629aa0fd1b1f58d388438710c (patch) | |
tree | 4385639c68d59b471a3dad3c433c1c596cd57f1b /Master/texmf-dist/doc/latex/eqexam/examples/quiz01.tex | |
parent | 399ec283a3bec858af23543f87ae19573572d207 (diff) |
rm eqexam, author request
git-svn-id: svn://tug.org/texlive/trunk@19295 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/eqexam/examples/quiz01.tex')
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diff --git a/Master/texmf-dist/doc/latex/eqexam/examples/quiz01.tex b/Master/texmf-dist/doc/latex/eqexam/examples/quiz01.tex deleted file mode 100644 index abedc62eab4..00000000000 --- a/Master/texmf-dist/doc/latex/eqexam/examples/quiz01.tex +++ /dev/null @@ -1,104 +0,0 @@ -\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} |