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\markright{Chapter 2: Answers 4}
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\Bf{Chapter 2: Answers 4 \hfill Jack K. Cohen  \hfill Colorado School of Mines}
\begin{enumerate}
\item (2.3.16) For $x < 7$, the quantity $x - 7$ is \Em{negative}, so $|x - 7| = -(x - 7) = 7 - x$.  Hence $\lim_{x \rightarrow 7^-}$ {\Large $\frac{7 - x}{|x - 7|}$} $ = \lim_{x \rightarrow 7^-}$ {\Large $\frac{7 - x}{7 - x}$} $ = 1$.  As a check:

 \Tt{LimitTable[(7 - x)/Abs[x - 7], \{x, 7, 6, -1\}] //TableForm} gives:
\begin{verbatim}
6.        1.
6.9       1.
6.99      1.
6.999     1.
6.9999    1.
6.99999   1.
\end{verbatim}
(This is our implementation of the recent ``Program of the Week".  The last two numerical arguments respectively give the number of table entries and the starting point offset from the limit point (so the table starts with 7 - 1 = 6 in the present case).

\item (2.3.20) $\sqrt{(4 + x)^2} = |4 + x|$. For $x < -4$ (e.g. $x = -5$), the quantity $4 + x$ is \Em{negative}, thus $|4 + x| = -(4 + x)$ and so the limit is -1.  Alternately,
\begin{verbatim}
LimitTable[(4 + x)/Sqrt [(4 + x)^2], {x, -4, 6, -1}] //TableForm
-5.        -1.
-4.1       -1.
-4.01      -1.
-4.001     -1.
-4.0001    -1.
-4.00001   -1.
\end{verbatim}

\item (2.3.26) $x = 5$ is the point where the limits don't exist.  The function becomes positive large no matter which side we approach 5 from: $\lim_{x \rightarrow 5^-} = \lim_{x \rightarrow 5^+} = +\infty$.


\end{enumerate}
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