\documentstyle[11pt,fleqn,epsf,calc]{article}
\markright{Chapter 2: Answers 2}
\begin{document}

\Bf{Chapter 2: Answers 2 \hfill Jack K. Cohen  \hfill Colorado School of Mines}
\begin{enumerate}
\item
\begin{verbatim}
f[x_] :=  x^3 + 3x^2
f[2]
    20
    
f[x] /. x -> 2
    20
    
f[t]
       2    3
    3 t  + t
    
 f[cat]
         2      3
    3 cat  + cat
    
Table[f[n], {n, 1, 5}]
    {4, 20, 54, 112, 200}
    
Plot[f[x], {x, -2, 2}]
    See Figure 1.
\end{verbatim}

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans2p1.eps}}
\caption{Plot of $x^3 + 3x^2$.} 
\end{figure}
\item  $\lim_{x \rightarrow 0}$ {\Large $\frac{\sqrt{x + 25} - 5}{x}$} is 1/10.
	\begin{enumerate}
	\item Figure 3  shows the output from 
	
	\Tt{Plot[(Sqrt[x + 25] - 5)/x, \{x, -2, 2\}, PlotRange -> \{0, .5\}]}

\begin{figure}[htb]
\epsfysize 110pt
\centerline{\epsffile{ans2p2.eps}}
\caption{Plot for problem 2.} 
\end{figure}

	\item Here is the output from
	
	\Tt{Table[\{x, (Sqrt[x + 25] - 5)/x\}, \{x, -1/5, 1/5, 1/20\}] //N //TableForm}
\begin{verbatim}
    -0.2    0.100201
    -0.15   0.10015
    -0.1    0.1001
    -0.05   0.10005
    0       Indeterminate
    0.05    0.09995
    0.1     0.0999002
    0.15    0.0998504
    0.2     0.0998008
\end{verbatim}
It may seem more natural to ``force'' \Mma\ to use decimal arithmetic during the calculation instead of evoking \Tt{N} afterwards, perhaps like this:

 \Tt{Table[\{x, (Sqrt[x + 25] - 5)/x\}, \{x, -.2, .2, .05\}] //TableForm}.
 
However, with in-exact arithmetic, the user has to be alert enough to reject output that is \Em{too} close to the singular point:
\begin{verbatim}
-0.2             0.100201
-0.15            0.10015
-0.1             0.1001
-0.05            0.10005
           -17
-1.38778 10      0.
0.05             0.09995
0.1              0.0999002
0.15             0.0998504
0.2              0.0998008
\end{verbatim}

	\item If Example 12 seems obscure, ask your instructor for help.
	\item  As indicated, $\sqrt{x + 25}  \approx 5 + x/10$, so  we can replace the given limit by $\lim_{x \rightarrow 0}$ {\Large $\frac{(5 + x/10) -5}{x}$} $ = 1/10$.
	\item Running the indicated code gives:
\begin{verbatim}
0.1       0.0999002
0.01      0.09999
0.001     0.099999
0.0001    0.0999999
0.00001   0.1
\end{verbatim}
With \Tt{x -> a - 1/10\^{}k}, we get:
\begin{verbatim}
-0.1       0.1001
-0.01      0.10001
-0.001     0.100001
-0.0001    0.1
-0.00001   0.1
\end{verbatim}

\noindent Note: We have supplied a function called \Tt{LimitTable} that uses this code as its core.

	\item Your opinion is more important than mine.
	\end{enumerate}

\item (2.2.8) $\lim_{t \rightarrow 2}$ {\Large $\frac{t^2 + 2t - 5}{t^3 - 2t}$} $ = 3/4$.
 
\item (2.2.22) $\lim_{x \rightarrow 9}$ {\Large $\frac{3 - \sqrt{x}}{9 - x}$} $ = 1/6$.

\item (2.2.32) $\lim_{x \rightarrow 1} $ {\Large $\frac{x^4 - 1}{x - 1}$} $ = 4$.

\item (2.2.35) $\lim_{x \rightarrow 0}$ {\Large $\frac{1}{x} \left( \frac{1}{x + 5} - \frac{1}{5} \right)$} $ = -1/10$.

\item (2.2.36) $\lim_{x \rightarrow 8}$ {\Large $\frac{x^{2/3} - 4}{x - 8}$} $ = 1/3$.
         To do this algebraically is a bit trickey.  Factor the numerator:  $(x^{1/3}  - 2)(x^{1/3}  + 2)$
	 and the denominator: $(x^{1/3}  - 2)(x^{2/3} + 2 x^{1/3} + 4)$.  Then cancel the common 
	 term and substitute $ x = 8$.
\end{enumerate}
\end{document}
