\documentclass{article} \usepackage[fleqn]{amsmath} \usepackage[pdf,cfg=quiz,forpaper,pointsonleft, % compile with exactly one of the following three nosolutions % answerkey % vspacewithsolns ]{eqexam} \examNum{1}\numVersions{2}\forVersion{a} \longTitleText {Quiz~\nExam--003} {Quiz~\nExam--007} \endlongTitleText \shortTitleText {Q{\nExam}s3} {Q{\nExam}s7} \endshortTitleText \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} \vspacewithkeyOn \solAtEndFormatting{\eqequesitemsep{3pt}} \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 labeled DNE (for ``does not exist''). \end{instructions} \begin{problem}[4] Compute $ \vA{\lim_{x\to-1}\frac{4x^2+x}{x}}\vB{\lim_{x\to2}\frac{1-3x}{x+1}}$ \begin{solution}[2in] As discussed in class, this is a ``Skill Level 0'' limit problem: \[ \begin{verA} \lim_{x\to-1}\frac{4x^2+x}{x} = \frac{4(-1)^2+(-1)}{-1} = \boxed{-3} \end{verA} \begin{verB} \lim_{x\to2}\frac{1-3x}{x+1} \lim_{x\to2}\frac{1-3(2)}{2+1} = \boxed{-\frac{5}{3}} \end{verB} \] \ifkeyalt\adjDisplayBelow\fi \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\vA{-2^-}\vB{-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} \ifkeyalt\adjDisplayBelow\fi \end{solution} \end{problem} \begin{problem}[3] Compute $\vA{\lim_{x\to2} \frac{1-x}{(x-2)^2}} \vB{\lim_{x\to3} \frac{x-2}{(3-x)^2}}$ \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} \ifkeyalt\adjDisplayBelow\fi \end{solution} \end{problem} \end{exam} \end{document}