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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 3B Solutions, September 29, 2017}
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\noindent
Section 2.1
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\noindent
2. Construct a table like the one given in Problem 1.  With $s(t) = 4.9t^2$, 
$$\begin{array}{ccccccccccccc}
\hline
{\rm Interval} &\vert& [3,3.1] &\vert& [3,3.01] &\vert& [3,3.001] &\vert& [3,3.0001] &\vert\\ \hline 
{\rm Ave. vel.} &\vert& 29.89 &\vert& 29.449 &\vert& 29.4049 &\vert& 29.4005&\vert  \\ \hline
\end{array}$$  
From this we guess the instantaneous velocity at $t = 3$ is about $29.4$.
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\noindent
40. 
\begin{enumerate}
\item[(a)]  The units of the rate of change are seconds/meter.  This measures
how fast the period is changing as the length of the pendulum increases.
\item[(b)]  The slope of line A (the tangent line) represents the instantaneous
rate of change of the period with respect to the length when the length is
$3$ meters.  The slope of line B (the secant line)  represents the average
rate of change of the period as a function of the length, over the interval
${\rm Length}$ in $[1,3]$.  
\item[(c)]  From the formula $T = \frac{3}{2} \sqrt{L}$ given in the problem,
we compute a table of values for the average rates of change
$$\begin{array}{ccccccccccccc}
\hline
{\rm Interval} &\vert& [3,3.1] &\vert& [3,3.01] &\vert& [3,3.001] &\vert& [3,3.0001] &\vert\\ \hline 
{\rm Ave.} &\vert& .4296 &\vert& .4327 &\vert& .4330 &\vert& .4330 &\vert \\ \hline
\end{array}$$  
It looks as though the limit is about $.4330$.  The average rates of change
over intervals ending at $L = 3$ would be tending to the same value.  
\end{enumerate}
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\noindent
Section 2.2
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\noindent
5.  It seems that $\lim_{x\to 0.5} f(x) = 1.5$ since the $y$-coordinates
are tending to that value as $x$ tends to $0.5$ from both sides.
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6.  It seems that $\lim_{x\to 0.5} f(x) = 1.5$ since the $y$-coordinates
are tending to that value as $x$ tends to $0.5$ from both sides.  Unlike 
the picture in 5, this function has $f(0.5) = 1 \ne 1.5$.  That does not
affect the limit.   
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\noindent
40.  
\begin{enumerate}
\item[] At $x=1$, $\lim_{x \to 1^-} f(x) = 3$ and $\lim_{x \to 1^+} f(x) = 1$.
This implies $\lim_{x \to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
\item[] At $x = 2$, the situation is similar. $\lim_{x \to 2^-} f(x) = 2$ and 
$\lim_{x \to 2^+} f(x) = 1$, so $\lim_{x \to 1} f(x)$ \emph{does not exist} 
(the one-sided limits are not equal).
\item[]  At $x = 4$ both one-sided limits exist and equal 2, so $\lim_{x\to 4} f(x) = 2$.
\end{enumerate}
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1^+} f(x) = 1$.
This implies $\lim_{x\to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
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