%% Math 131 Midterm 1, Fall 2004
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\textbf{College of the Holy Cross, Fall Semester, 2015} \\
\textbf{MATH 133, section 1, Solutions for Midterm 2} \\
\textbf{Thursday, October 22}
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\vskip 10pt
\begin{enumerate}

\item{} An object moves along a straight line path with position given by 
$x(t) = 2t^3 + t$, ($t$ in seconds, $x$ in feet).   
\begin{enumerate}
\item[(a)] (5) What is the average velocity of the object on the interval $[1,4]$?
\item[ ] {\it Solution:}  The average velocity is 
$$v_{ave} = \frac{x(4) - x(1)}{4 - 1} = \frac{132 - 3}{4 - 1} = 43\ {\rm ft/sec}$$
\item[(b)] (10)  Fill in the following table with average velocities computed over the indicated intervals.  Using this information, estimate the \emph{instantaneous velocity} at $t = 1$.  
$$\begin{array}{ccccccccc}
\hline
{\rm interval} &\vline& \qquad[1,2]\qquad &\vline& \quad [1,1.1]\quad &\vline& \quad [1,1.01]\quad &\vline& \quad [1,1.001]\quad \\ \hline
{\rm ave. vel.}&\vline& 15 &\vline&  7.62  &\vline&  7.0602  &\vline&  7.006002 \\ \hline 
\end{array}$$
\item[ ] Estimated instantaneous velocity $=$ probably 7 ft/sec.
\end{enumerate}

\item{} Answer all parts of this question by referring to the graph $y = f(x)$ [on next page].
\begin{enumerate}
\item[(a)] (6) $\displaystyle{\lim_{x\to 0^-} f(x)} = 3$ and 
$\displaystyle{\lim_{x\to 0^+} f(x)} = 1$.
\item[(b)] (6) $f(x)$ has an infinite discontinuity at $x = 1$ (one-sided infinite
discontinuities are possible).
\item[(c)] (3)  True/False:  The limit $\displaystyle{\lim_{x\to -1} f(x)}$ does not exist.  
False -- we have $\displaystyle{\lim_{x\to -1} f(x) = 4}$.  The value $f(-1) = 3$ does
not affect the limit.  
\item[(d)] (9)  $\displaystyle{\lim_{x\to 2^-} f(x)} = 1$ and 
$\displaystyle{\lim_{x\to 2^+} f(x)} = 1$.  
Given that $f(2) = 1$, what can we
say about $f$ at $x = 2$?  $f$ is \emph{continuous} at $x = 2$.  
\item[(e)] (6)  True/False:  $f(x)$ has a removable discontinuity shown 
in this part of the graph.  True.  If so, where is it? 
$x = -1$.  The limit exists, but is different from $f(-1)$.  That means 
$f$ has removable discontinuity there.    
\end{enumerate} 
\begin{figure}
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\includegraphics[width=.5\linewidth]{FPlot.png}
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\item{}  Compute \emph{any four} of the following limits. (Only the best four will be counted
for your total score.)
\begin{enumerate}
\item[(a)] (10) This is zero/zero as $x \to 3$, so we try to factor and cancel, then
compute the limit by continuity:
\begin{eqnarray*}
\lim_{x\to 3} \frac{x^2 - 5x + 6}{3x^2 - 9x}
&=& \lim_{x\to 3} \frac{(x - 2)(x - 3)}{3x(x - 3)}\\
&=& \lim_{x\to 3} \frac{x-2}{3x}\\
&=& \frac{1}{9}.
\end{eqnarray*}
\item[(b)] (10) This is a zero/zero form with a difference of square roots, so
we use the conjugate radical:
\begin{eqnarray*}
\lim_{h \to 4} \frac{\sqrt{12 + h} - \sqrt{16}}{h^2-16}
&=& \lim_{h\to 4} \frac{(\sqrt{12 + h} - \sqrt{16})(\sqrt{12 + h} + \sqrt{16})}
{(h^2 - 16)(\sqrt{12 + h} + \sqrt{16})}\\
&=& \lim_{h\to 4}\frac{12 + h - 16}{(h-4)(h+4)(\sqrt{12 + h} + \sqrt{16})}\\
&=& \lim_{h\to 4}\frac{h - 4}{(h-4)(h+4)(\sqrt{12 + h} + \sqrt{16})}\\
&=& \lim_{h\to 4}\frac{1}{(h+4)(\sqrt{12 + h} + \sqrt{16})}\\
&=& \frac{1}{64}.
\end{eqnarray*}
\item[(c)]  (10) Use the Limit Product Law and $\lim_{t\to 0} \frac{\sin(t)}{t} = 1$:
$$\lim_{t\to 0} \frac{(t^2 + 5)\sin(t)}{t} = \lim_{t\to 0} (t^2 + 5) \lim_{t\to 0} \frac{\sin(t)}{t} = 5\cdot 1 = 5.$$ 
\item[(d)]  (10) For the limit as $x \to \infty$, multiply the top and bottom by 
$\frac{1}{x^3}$:
\begin{eqnarray*}
\lim_{x\to \infty} \frac{x^3 + 3x + 1}{7x^3 + x^2 + 4x} &=& 
\lim_{x\to \infty} \frac{(x^3 + 3x + 1)\frac{1}{x^3}}{(7x^3 + x^2 + 4x)\frac{1}{x^3}}\\
&=&\lim_{x\to \infty} \frac{1 + \frac{3}{x^2} + \frac{1}{x^3}}{7 + \frac{1}{x} + \frac{4}{x^2}}\\
&=&\frac{1}{7}.
\end{eqnarray*}
\item[(e)] (10)  Put the two terms on top over a common denominator, simplify
the fraction, then factor and cancel to take the limit: 
\begin{eqnarray*}
\lim_{x\to 2} \frac{\frac{1}{x^2} - \frac{1}{4}}{x - 2} 
&=& \lim_{x\to 2} \frac{4 - x^2}{4x^2(x - 2)}\\
&=& \lim_{x\to 2} \frac{-(x + 2)(x - 2)}{4x^2(x - 2)}\\
&=& \lim_{x\to 2} \frac{-(x + 2)}{4x^2}\\
&=& \frac{-1}{4}.
\end{eqnarray*}
\end{enumerate}


\item{} Let $f(x) = x^2 + 4x + 1$.  
\begin{enumerate}
\item[(a)] (5)  What is the slope of the secant line to the graph through the 
points $(1,6)$ and $(2,13)$?
\item[ ] {\it Solution:}  The slope is
$$\frac{13 - 6}{2 - 1} = 7.$$
\item[(b)] (5)  Give a general formula for the slope of the secant line
through the points $(1,6)$ and $(1+h,(1+h)^2 + 4(1 + h) + 1)$.
\item[ ] {\it Solution:}  The general secant line slope is 
$$m_{sec} = \frac{(1 + h)^2 + 4(1 + h) + 1 - 6}{(1+h) - 1} = \frac{1 + 2h + h^2 + 4h + 4 + 1 - 6}{h}
= \frac{6h + h^2}{h}.$$
\item[(c)] (5)  Find the limit as $h \to 0$ of your slope from 
part (b).
\item[] {\it Solution:}  Again a zero/zero form.  We can cancel $h$'s and then
use continuity:
$$\lim_{h\to 0} \frac{6h + h^2}{h} = \lim_{h\to 0} \frac{h(6 + h)}{h} = \lim_{h\to 0} 6+ h = 6.$$

\end{enumerate}
\end{enumerate}

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