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\markright{Math 135, Section 1, Midterm 2}

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\begin{document}

\begin{center}
\textbf{College of the Holy Cross}\\
\textbf{Math 135, Section 1 -- Solutions for Midterm Exam 2}\\
\textbf{Friday, November 1}
\end{center}

\vskip 10pt
\begin{enumerate}

\item[I.] Compute the indicated limits. You must show all necessary work to justify your answer to 
receive full credit.

  \begin{enumerate}

  \item
    (5) $\ds\lim_{x\to 1}\frac{x^2 - 5x + 2}{2x^2 - 2x + 3}$
\item[] {\it Solution:}  The top is going to $-2$ and the bottom is going to 
$3$ as $x\to 1$.  By the Limit Quotient Rule, the limit is $= -2/3$.  
  \item 
    (5) $\ds\lim_{x\to 1}\frac{x^2 - 7x + 6}{2x^2 - 2x}$
\item[] {\it Solution:}  This is a $0/0$ indeterminate limit so we should factor the 
top and bottom and try to cancel:
\begin{eqnarray*}
\lim_{x\to 1}\frac{x^2 - 7x + 6}{2x^2 - 2x} &=& \lim_{x\to 1} \frac{(x-6)(x-1)}{2x(x-1)}\\
&=& \lim_{x\to 1} \frac{x - 6}{2x}\\
&=& -5/2.
\end{eqnarray*}
  \item 
    (5) $\ds\lim_{x\to\infty}\frac{x^2 - 5x + 2}{2x^2 - 2x + 3}$
\item[] {\it Solution:} Divide the top and bottom by $x^2$ then take the limit as $x \to \infty$:
\begin{eqnarray*}
\lim_{x\to \infty} \frac{x^2 - 5x + 2}{2x^2 - 2x + 3} &=&
\lim_{x\to \infty} \frac{1 -5/x + 2/x^2}{2 - 2/x + 3/x^2}\\
&=& 1/2.
\end{eqnarray*}
  \item 
    (5) $\ds\lim_{\theta\to 0} \frac{\sin(4\theta)}{\sin(5\theta)}$
\item[] {\it Solution:}  Since $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$, this limit
equals 
$$\lim_{\theta\to 0} \frac{4}{5} \frac{\frac{\sin(4\theta)}{4\theta}}{\frac{\sin(5\theta)}{5\theta}}
= \frac{4}{5}.$$
\end{enumerate}

\item{} The graph of the function 
$$f(x) = \begin{cases}
               \frac{1}{x+1} & x < -1\\
               -x + 2 & -1 \le x \le 0\\
               \frac{4x^3 - 24x^2 + 44x - 24}{x^3 - 5x^2 + 2x + 8} & 0 < x < 3
\end{cases}$$
 is shown in Figure 1. 
\begin{figure}  
  \begin{center}
	\includegraphics[scale = .3]{M2II.png}
  \end{center}
\end{figure}

\begin{enumerate}
  \item (10) What are $\ds\lim_{x\to 0^-} f(x)$ and $\ds\lim_{x\to 0^+} f(x)$?  (In your answer say clearly which is which.)

\item[] {\it Solution:} $\lim_{x\to 0^-} f(x) = \lim_{x \to 0^-} -x + 2 = 2$ and 
$$\lim_{x\to 0^+} = \lim_{x\to 0^+} \frac{4x^3 - 24x^2 + 44x - 24}{x^3 - 5x^2 + 2x + 8} = -3$$
(These can also be estimated from the graph.

  \item (15) Find all $x$ in $(-2,3)$ where $f$ is discontinuous. Give the types of each of the discontinuities.
\item[] {\it Solution:}  There are {\it three}:  $x = -1$ is an {\it infinite discontinuity} since
$\lim_{x\to -1^-} \frac{1}{x + 1} = -\infty$. $x = 0$ is a {\it jump discontinuity} since the 
one-sided limits exist but are unequal.  Finally, 
$x = 2$ is a {\it removable discontinuity}.  Note the open circle at $x = 2$ in the graph.
This is a reflection of the fact that the function 
$$\frac{4x^3 - 24x^2 + 44x - 24}{x^3 - 5x^2 + 2x + 8}$$
gives $0/0$ at $x = 2$, but the limit as $x \to 2$ does exist and equals $2/3$.  
\end{enumerate}

\item{} Do not use the short-cut differentiation rules from Chapter~3 in this
  question.

  \begin{enumerate}

  \item (5) State the limit definition of the
  derivative~$f'(x)$.
  \item[] {\it Solution:}
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$
(provided that the limit exists).  

  \item (10) Estimate the derivative of $f(x) = \ds\sqrt{x + 2}$ at $a = 7$ \emph{numerically}
  by computing difference quotients of $f$ with $h = \pm .1$, then $h = \pm .01$.  Enter 
  your values in the table below, and then state what your estimate of $f'(7)$ is.  
\item[] {\it Solution:}
$$\begin{array}{ccccccccccc}
\hline
h &\vline& -.1 \qquad &\vline& -.01 \qquad &\vline& .01 \qquad &\vline& .1 \qquad&\vline\\ \hline
{\rm difference\ quotient\ value}&\vline&    .1671  &\vline& .1667    &\vline& .1666 &\vline& .1662  &\vline \\ \hline 
\end{array}$$
\vskip 10pt
\noindent
$f'(6) \doteq \underline{.1666}$

  \item (10) Use the definition to
    compute the derivative function of $f(x)=\ds\sqrt{x + 2}$.

\item[] {\it Solution:}  Multiply by the conjugate radical, simplify, and take the limit:
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{\sqrt{x+h+2} - \sqrt{x+2}}{h}\\
&=& \lim_{h to 0} \left(\frac{\sqrt{x+h+2} - \sqrt{x+2}}{h}\right)
\cdot\left(\frac{\sqrt{x+h+2} + \sqrt{x+2}}{\sqrt{x+h+2} + \sqrt{x+2}}\right)\\
&=& \lim_{h\to 0} \frac{(x+h+2) - (x+2)}{h(\sqrt{x + h+2} + \sqrt{x+2})}\\
&=& \lim_{h\to 0} \frac{h}{h(\sqrt{x + h+2} + \sqrt{x+2})}\\
&=& \lim_{h\to 0} \frac{1}{\sqrt{x + h+2} + \sqrt{x+2}}\\
&=& \frac{1}{2\sqrt{x+2}}
\end{eqnarray*}
  \item (5) Find the equation of the line tangent to
    the graph $y=\ds\sqrt{x + 2}$ at $a=7$.
\item[] {\it Solution:}  The point on the graph is $(7,3)$.
The slope of the tangent line is $f'(7) = \frac{1}{6}$ (You could also use the estimated
value $f'(x) \doteq .1666$ from part b.)  The equation of the tangent line is 
$$y - 3 = \frac{1}{6}(x - 7),$$
or 
$$y = \frac{1}{6}x + \frac{11}{6}$$
in slope-intercept form.
  \end{enumerate}


\vfil\eject
\item{} Use the short-cut 
  rules to compute the following derivatives. You may use any correct
  method, but you must show work for full credit.

\begin{enumerate}
  \item (5) $\ds\frac{d}{dx}\left(\frac{3}{\sqrt{x}} - e^x + 3x\right)$
\item {\it Solution:}  First rewrite the function using rules for exponents as 
$$3x^{-1/2} - e^x + 3x$$
Then the derivative is 
$$\frac{-3}{2}x^{-3/2} - e^x + 3$$

  \item (10) $\ds\frac{d}{dv}\left((v^2 - 2v)(v^3 + 1)\right)$
\item {\it Solution:}  By the product rule, the derivative is 
$$(v^2 - 2v)(3v^2) + (v^3+1)(2v - 2).$$
This form is OK by the directions.  The simplified form is
$$5v^4 - 8v^3 + 2v - 2.$$
(Note:  This could also be done by multiplying out the product, then differentiating
term by term using the power rule.)

  \item (10) $\ds\frac{d}{dx}\left(\frac{x^2 - 4x + 3}{x^4 - x}\right)$
\item {\it Solution:}  By the quotient rule, the derivative is
$$\frac{(x^4 - x)(2x - 4) - (x^2 - 4x + 3)(4x^3 -1)}{(x^4-x)^2}.$$
Again, by the directions, this form is OK; the simplified form
is 
$$\frac{-2x^5 + 12x^4 - 12 x^3 - x^2 + 3}{(x^4 - x)^2}.$$
\end{enumerate}
\end{enumerate}
\end{document}
