\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \subsection{MATC 1200, Calculus for Business and Social Sciences} \subsection{Lab 2, Week of April 6, 2009} \subsection{Limits of Functions\protect\bigskip} In this lab we will study the behavior of a function near a specified point. \ While this is sometimes a straightforward process, it can also be quite subtle. \ In many instances in calculus the process for finding a limit must be applied carefully. \ **Before you start\ this lab do the following: \ Select \textbf{Tools/Computation Setup/General }and change the \textbf{% Digits Shown in Results} to $8$.**\medskip \begin{enumerate} \item Consider the function $f$ defined by $f(x)=\dfrac{x^{3}-8}{x-2}$. \ Type this equation and define the function $f$ in SN. \ (If your fraction appears too small you can make it larger by highlighting it and selecting the \FRAME{itbpF}{22.4375pt}{21.0625pt}{2pt}{}{}{property.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 22.4375pt;height 21.0625pt;depth 2pt;original-width 17.8125pt;original-height 16.9375pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/property.wmf';file-properties "XNPEU";}}button on your toolbar. \ Then choose \textbf{Big.})\medskip \begin{enumerate} \item What is the domain of $f(x)?$ $\ $Explain.\medskip \item Use SN to graph the function $f$ and change the viewing window to $% -4\leq x\leq 6$, and $-2\leq y\leq 55$.\medskip \item Modify the graph in \textsf{1b} to reflect the fact that the domain of $f(x)$ is all real numbers except for one value.\medskip \item Use SN to evaluate the function $f$ at $x=3.8,$ $3.9,$ $3.99,$ $3.999$ and $3.9999$. \ Record your answers in the table.\medskip $\ \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $3.8$ & $3.9$ & $3.99$ & $3.999$ & $3.9999$ \\ \hline $f(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% $\medskip \item Do a similar experiment on $f$ for values of $x$ slightly greater than $4$. \ Find the values of $f(x)$ at $x=4.2,$ $4.1,$ $4.01,$ $4.001$ and $% 4.0001$. \ Again, record your results in the table.\medskip $% \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $4.2$ & $4.1$ & $4.01$ & $4.001$ & $4.0001$ \\ \hline $f(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% $\medskip \item Based on your answers in parts \textsf{1d} and \textsf{1e, }what is $% \lim\limits_{x\rightarrow 4}f\left( x\right) $? \ Explain your answer.\medskip \item By hand, evaluate $f(4)$. \ Show the calculation below.\medskip Note: You should find that $\lim\limits_{x\rightarrow 4}f\left( x\right) =f(4)$ so we say the function $f(x)$ \textit{is continuous at }$x=4$. $\ $% Graphically, this means that the graph of the function has no breaks at $x=4$% .\medskip \item By hand, "algebraically" compute $\lim\limits_{x\rightarrow 4}\dfrac{% x^{3}-8}{x-2}$. \ Show your computation below.\medskip \end{enumerate} \item Use the same function $f(x)=\dfrac{x^{3}-8}{x-2}$ as above, but this time consider what happens as $x$ approaches $2$.\medskip \begin{enumerate} \item Study the behavior of $f\left( x\right) $ near $x=2$ as you did above by using SN to evaluate $f$ at $1.8,$ $1.9,$ $1.99,$ $1.999,$ $1.9999$ and at $2.2,2.1,2.01,$ $2.001,2.0001$. \ Record your results in the tables below.\medskip \[ \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $1.8$ & $1.9$ & $1.99$ & $1.999$ & $1.9999$ \\ \hline $f(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% \]% \medskip \[ \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $2.2$ & $2.1$ & $2.01$ & $2.001$ & $2.0001$ \\ \hline $f(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% \]% \medskip \item Based on your results in part \textsf{2a}, determine $% \lim\limits_{x\rightarrow 2}f(x)$. \ Explain your answer.\medskip \item Can you evaluate $f(2)$? \ Explain your answer.\medskip Note: \ You should find that $\lim\limits_{x\rightarrow 2}f(x)\neq $ $f(2).$ In this case, we say $f(x)$ is \textit{discontinuous at }$x=2$. \ Graphically, this means the graph of $f(x)$ will have a break of some form at $x=2$.\medskip \item To evaluate $\lim\limits_{x\rightarrow 2}\dfrac{x^{3}-8}{x-2}$ algebraically we cannot simply evaluate $f(2).$ \ Notice that if you try to evaluate $\dfrac{x^{3}-8}{x-2}$ at $x=2$ you get $\dfrac{2^{3}-8}{2-2}=% \dfrac{0}{0}$. \ We have discussed in lecture that this is an \textsf{% indeterminate form }for limits. \ As a first step towards evaluating the limit, factor $x^{3}-8$ using SN.\medskip \item Now, use this factorization to simplify the expression\textbf{\ }$% \dfrac{x^{3}-8}{x-2}$ by hand. \ You should have a quadratic expression left over.\medskip \item Use your work in part e to "algebraically" compute $% \lim\limits_{x\rightarrow 2}\dfrac{x^{3}-8}{x-2}$ by hand. \ Show your computation below.\medskip You should note that whenever we have a limit with indeterminate form $% \dfrac{0}{0}$ we will need to do some algebraic simplification before we can evaluate the limit.\medskip \item Let's step back for a moment and reflect on our method. \ Let $g\left( x\right) $ be the new (quadratic) function you found in \textsf{2e}. \ What is the domain of the simplified function $g\left( x\right) $? \ How is this different from the domain of $f(x)$?\medskip \item The graphs of $g\left( x\right) $ and $f\left( x\right) $ produced in SciNotebook will appear to be identical.\ \ How are the graphs of $g(x)$ and $f(x)$ actually different?\medskip You should notice that $g(x)=f(x)$ for all values of $x$ except for $x=2$. \ Thus, we can conclude that $\lim\limits_{x\rightarrow 2}g(x)$ will be the same as $\lim\limits_{x\rightarrow 2}f(x)$. \ **The limit question asks for the behavior of the functions for values of $x$ near $2$ (but not equal to $% 2 $).**\medskip \end{enumerate} \item Consider the function $h(x)=\dfrac{\allowbreak 2x^{3}+2x^{2}-5x+21}{x+3% }$.\medskip \begin{enumerate} \item By hand, find the $\lim\limits_{x\rightarrow \,-3}(2x^{3}+2x^{2}-5x+21) $ and $\lim\limits_{x\rightarrow \,-3}\left( x+3\right) .$ \ (Hint: Both of these functions are polynomials and can be evaluated directly.)\medskip \item Based solely on your answer to part \textsf{a}, what can you say about $\lim\limits_{x\rightarrow \,-3}h(x)$?\medskip \item In two or three words, what is the next step you should take in order to find $\lim\limits_{x\rightarrow \,-3}h(x)$?\medskip \item Take that next step and find $\lim\limits_{x\rightarrow \,-3}h(x)$. \ Show all of the work you do and the results obtained from SN below.\medskip \end{enumerate} \item The notation $\lim\limits_{x\rightarrow \infty }k(x)$ (or $% \lim\limits_{x\rightarrow \,-\infty }k(x)$) is used to denote the limit of a function at infinity. \ For $\lim\limits_{x\rightarrow \infty }k(x)$ we consider what happens to the outputs of $k(x)$ as $x$ approaches infinity. \ By comparison, for $\lim\limits_{x\rightarrow \,-\infty }k(x)$ we consider what happens to the outputs of $k(x)$ as $x$ approaches minus infinity.\medskip Consider the function $k(x)=\dfrac{-3x}{\sqrt{10+x^{2}}}$.\medskip \begin{enumerate} \item To find $\lim\limits_{x\rightarrow \infty }k(x)$ numerically, use SN\ to evaluate $k(x)$ numerically at six values of $x$ that are approaching infinity. \ Record your answers in the table.\medskip $% \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $10$ & $100$ & $1000$ & $10000$ & $100000$ \\ \hline $k(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% $\medskip \item Use your results to make a conjecture about the value of $% \lim\limits_{x\rightarrow \infty }k(x)$.\medskip \item To find $\lim\limits_{x\rightarrow \text{ }-\infty }k(x),$ use SN to evaluate $k(x)$ numerically at six very small values of $x$ that are approaching minus infinity. \ Record your answers in the table.\medskip $% \begin{tabular}{|l||l|l|l|l|l|} \hline $x$ & $-10$ & $-100$ & $-1000$ & $-10000$ & $-100000$ \\ \hline $k(x)$ & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} & \hspace{1.1in} \\ \hline \end{tabular}% $\medskip \item Use your results to make a conjecture about the value of $% \lim\limits_{x\rightarrow \,-\infty }k(x)$.\medskip \item Sketch a graph of $k(x)$ for the given window.\medskip \FRAME{dtbpFX}{% 3.4791in}{2.3194in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.4791in;height 2.3194in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-10";xmax "10";xviewmin "-10";xviewmax "10";yviewmin "-5";yviewmax "5";viewset"XY";rangeset"X";plottype 4;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 100;plotstyle "patch";axesstyle "normal";axestips FALSE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$0$};linecolor 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