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\markright{MATH 133, section 1, Midterm 2}

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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2017} \\
\textbf{MATH 133, Midterm 2 Solutions} \\
\textbf{Friday, October 20}
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\begin{enumerate}

\item{} An object moves along a straight line path with position given by 
$x(t) = 4t^2 + 6t$, ($t$ in seconds, $x$ in feet).   
\begin{enumerate}
\item[(a)] What is the average velocity of the object on the interval $[0,3]$?
\item[] {\it Solution:}  The average velocity is 
$$v_{ave} = \frac{x(3) - x(0)}{3 - 0} = \frac{4\cdot 3^2 + 6\cdot 3 - 0}{3 - 0} = 18$$
(units are feet per second).
\item[(b)] Fill in the following table with average velocities computed over the indicated intervals.  Using this information, estimate the \emph{instantaneous velocity} at $t = 0$.  
\item[]{\it Solution:}
$$\begin{array}{ccccccccc}
\hline
{\rm interval} &\vline& \qquad[0,1]\qquad &\vline& \quad [0,0.1]\quad &\vline& \quad [0,0.01]\quad &\vline& \quad [0,0.001]\quad \\ \hline
{\rm ave. vel.}&\vline&    10  &\vline&   6.4    &\vline&  6.04      &\vline&  6.004    \\ \hline 
\end{array}$$
\item[ ] Estimated instantaneous velocity $= 6$.
\end{enumerate}
\item{} Answer all parts of this question by referring to the graph $y = f(x)$ in Figure 1.


\begin{enumerate}
\item[(a)] (5) $\displaystyle{\lim_{x\to 1^-} f(x)} = 2 $ and 
$\displaystyle{\lim_{x\to 1^+} f(x)} = 2$.
\item[(b)] (5) $f(x)$ has a jump discontinuity at $x = 3$.
\item[(c)] (5)  True/False:  The limit $\displaystyle{\lim_{x\to 2} f(x)}$ does not exist.  
False.  If so, what is the limit? If not, say why not: Limit is apparently $-1$.
\item[(d)] (5)  True/False:  $f(x)$ has an infinite discontinuity in this part of the graph.
False. 
$x = \underline{\qquad\qquad\qquad}$.  (If not leave this space blank.)  
\item[(e)] (5)  True/False:  $f(x)$ has a removable discontinuity shown 
in this part of the graph.  True.  If so, where is it? 
$x = 1$. 
\end{enumerate}

\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) 
$$\lim_{x\to 1} \frac{x^2 + 7x - 8}{x^2 - 1}$$
\item[] {\it Solution:}  This is a 0/0 limit.  We can factor, cancel, and evaluate as follows:
\begin{align*}
\lim_{x\to 1} \frac{x^2 + 7x - 8}{x^2 - 1}
&= \lim_{x\to 1} \frac{(x - 1)(x + 8)}{(x - 1)(x + 1)}\\
&= \lim_{x\to 1} \frac{(x + 8)}{(x + 1)}\\
&= 9/2.
\end{align*}

\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{Q3G.png}
\caption{$y = f(x)$ for problem 2.}
\end{center}
\end{figure}

\item[(b)] (10) 
$$\lim_{h \to 3} \frac{\sqrt{11 + h} - \sqrt{14}}{h-3}$$
\item[] {\it Solution:}  Also a $0/0$ indeterminate form.
We multiply top and bottom by the conjugate radical, simplify, and then substitute:
\begin{align*}
\lim_{h \to 3} \frac{\sqrt{11 + h} - \sqrt{14}}{h-3}
&= \lim_{h \to 3} \frac{(\sqrt{11 + h} - \sqrt{14})}{h-3}\cdot \frac{(\sqrt{11+h} + \sqrt{14})}{(\sqrt{11+h} + \sqrt{14})}\\
&= \lim_{h \to 3} \frac{11 + h - 14}{(h-3) (\sqrt{11+h} + \sqrt{14})}\\
&= \lim_{h \to 3} \frac{h - 3}{(h-3) (\sqrt{11+h} + \sqrt{14})}\\
&= \lim_{h \to 3} \frac{1}{(\sqrt{11+h} + \sqrt{14})}\\
&= \frac{1}{2\sqrt{14}}.
\end{align*}
\item[(c)]  (10) 
$$\lim_{t\to 0} \frac{t^2 + 5}{t + 1}.$$ 
\item[] {\it Solution:}  This one is not indeterminate -- it can be evaluated directly
using the limit quotient, sum, and product rules
$$\lim_{t\to 0} \frac{t^2 + 5}{t + 1} =\frac{0^2 + 5}{0 + 1} = 5.$$ 
\item[(d)]  (10) 
$$\lim_{x\to \infty} \frac{x^3 + 3x + 1}{7x^3 + x^2 + 4x}.$$
\item[] {\it Solution:}  We can multiply the top and bottom by $\frac{1}{x^3}$ to obtain:
\begin{align*}
\lim{x\to\infty} \frac{(x^3 + 3x + 1)\cdot \frac{1}{x^3}}{(7x^3 + x^2 + 4x)\cdot \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{align*} 
\item[(e)] (10) 
$$\lim_{t\to 0} \frac{\sin(9t)}{t}.$$ 
\item[] {\it Solution:}  For this one, we want to apply the formula 
$$\lim_{x\to 0} \frac{\sin(x)}{x} = 1$$
but that requires having exactly the same thing ``inside'' the sine function
and in the denominator.  If we multiply top and bottom by $9$ we get
$$\lim_{t\to 0} \frac{\sin(9t)}{t} = \lim_{t\to 0} \frac{9\sin(9t)}{9t} = 9\cdot 1 = 9.$$
\end{enumerate}

\item{} Let $f(x) = x^2 - 2x + 2$.  
\begin{enumerate}
\item[(a)] (5)  What is the slope of the secant line to the graph through the 
points $(1,1)$ and $(3,5)$?
\item[] {\it Solution:}
The slope is 
$$m_{sec} = \frac{5-1}{3-1} = 2.$$
\item[(b)] (5)  Give a general formula for the slope of the secant line
through the points $(1,1)$ and $(1+h,(1+h)^2 - 2(1 + h) + 2)$.
\item[] {\it Solution:}  The slope of the secant line is 
$$\frac{(1+h)^2 - 2(1 + h) + 2 - 1}{(1+h) - 1} = \frac{(1 + 2h + h^2) - 2 - 2h + 2 - 1}{h} = 
\frac{h^2}{h} = h$$
\item[(c)] (5)  Find the limit as $h \to 0$ of your slope from 
part (b).
\item[] {\it Solution:}  The limit is $0$.  
\item[(d)] (5)  What does your answer in part (c) tell you in terms of the graph
$y = f(x)$, related lines, etc.?  
\item[] {\it Solution:}  It is the slope of the tangent line to the graph $y = x^2 - 2x + 2$ at
$(1,1)$.  The slope is $0$ which means the tangent line is the horizontal line $y = 1$.  
\end{enumerate}
\end{enumerate}

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