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

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\begin{document}




\begin{enumerate}

\item{} Compute the indicated limits. Show all work for full credit.

  \begin{enumerate}

  \item[(a)]
    (5) $\ds\lim_{x\to 1}\frac{x^2 - 6x - 8}{2x^2 - 2x - 4}$
\item[] {\it Solution:}  This is a rational function, so it is continuous at every $x$
where the denominator is not zero.  Since $2\cdot (1)^2 - 2\cdot 1 - 4 = -4 \ne 0$ that
is true here.  The limit is the same as the value of the function at 1:  $= \frac{-13}{-4} = \frac{13}{4}$.  Note:  the same conclusion can be reached by using the limit sum, product, and quotient rules.
\end{eqnarray*}
\item[(b)]
    (5) $\ds\lim_{x\to 2}\frac{x^2 - 6x - 8}{2x^2 - 2x - 4}$
\item[] {\it Solution:}  This is a $0/0$ indeterminate form.  We factor the top and bottom
and cancel to obtain
$$\frac{x^2 - 6x - 8}{2x^2 - 2x - 4} = \frac{(x - 2)(x - 4)}{(x - 2)(2x + 2)} = \frac{x - 4}{2x+2}$$
for all $x \ne 2$.  This function is continuous at $x = 2$, and the limit is $\frac{-2}{6} = \frac{-1}{3}$.
  \item[(c)] 
    (5) $\ds\lim_{x\to\infty}\frac{x^2 - 6x - 8}{2x^2 - 2x - 4}$
\item[] {\it Solution:}  The limit is $\frac{1}[2}$, as can be seen by this calculation (multiply top and bottom
by $\frac{1}{x^2}$):
\begin{eqnarray*}
\lim_{x\to\infty}\frac{x^2 - 6x - 8}{2x^2 - 2x - 4} &=& \lim_{x\to\infty}\frac{1 - \frac{6}{x} - \frac{8}{x^2}}{2 - \frac{2}{x} - \frac{4}{x^2}}\\
&=& \frac{1}{2}.
\item[(d)] 
(5) $\ds \lim_{\theta\to 0} \frac{\sin(3\theta)}{\theta}$
\item[] {\it Solution:}
This equals
$$\lim_{\theta\to 0} 3 \frac{\sin(3\theta)}{3\theta} = 3\cdot 1 = 3.$$
\end{eqnarray*}
\end{enumerate}

\item{} The graph of a function~$f$ with $f(-1) = -.2$ and $f(2) = -1$ is shown below.  
  \begin{center}
	\includegraphics[scale = .4]{M2I.png}
  \end{center}

\begin{enumerate}
  \item[(a)] (10) What are $\ds\lim_{x\to -1^-} f(x)$ and $\ds\lim_{x\to -1^+} f(x)$?
\item[] {\it Solution:}  From the graph and the given information, 
$\ds\lim_{x\to -1^-} f(x) = -.2$ and $\ds\lim_{x \to -1^+} f(x) = 2$.
 \item[(b)] (15) Find all $x$ in $(-3,5)$ where $f$ is discontinuous. Give the types of each of the
discontinuities
\item[] {\it Solution:}  $f(x)$ has jump discontinuities at $x = -1$ and $x = 2$.  It also has an
infinite discontinuity (vertical asymptote) at $x = 3$.  These are the only discontinuities. 
  \item[(c)] (10) Given that $f(x) = x + 3$ for $-1 < x < 0$ and $f(x) = 3 - x - \frac{x^3}{8}$ for $0 \le x < 2$, is
$f$ differentiable at $a = 0$?  Why or why not? 
\item[] {\it Solution:}  The answer is \emph{no}, $f$ is not differentiable at $0$.  The easiest
way to see this is that $\ds\lim_{h\to 0^-} \frac{f(0+h) - f(0)}{h}$ will agree with the derivative of 
$x + 3$ at $x = 0$, and equal 1.  On the other hand $\ds\lim_{h\to 0^+} \frac{f(0+h) - f(0)}{h}$ will 
agree with the derivative of $3 - x - \frac{x^3}{8}$ at $x = 0$, which is $-1$.  (This is the precise
meaning of the apparent ``corner'' on the graph at $x = 0$.)  Since the one-sided limits
of the difference quotient of $f$ are not the same, $f'(0)$ does not exist. 

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

  \begin{enumerate}

  \item[(a)] (5) State the limit definition of the
  derivative~$f'(x)$.
\item[] {\it Solution:}  The derivative $f'(x)$ is 
$$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h},$$
if the limit exists.
\item[(b)]   Estimate the derivative of $f(x) = \ds\sqrt{x + 3}$ at $a = 6$ 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 $f'(8)$ is.  
\item[] {\it Solution:}  Computing 
$$\frac{\sqrt{6 + h + 3} - \sqrt{6+3}}{h} = \frac{\sqrt{9+h} - 3}{h}$$ for the given
values of $h$ we get the numbers in the table (rounded to 4 decimals):
$$\begin{array}{ccccccccccc}
\hline
h &\vline& -.1 \qquad &\vline& -.01 \qquad &\vline& .01 \qquad &\vline& .1 \qquad&\vline\\ \hline
{\rm difference\ quotient\ value}&\vline&    .1671     &\vline&   .1667    &\vline&  .1666      &\vline&  .1662 &\vline \\ \hline 
\end{array}$$
The limit is apparently around $.1666$.

  \item[(b)] (10) Use the definition to
    compute the derivative function of $f(x)=\ds\sqrt{x + 3}$.
\item[] {\it Solution:}  We compute the limit by multiplying top and bottom of the 
difference quotient by the conjugate radical:
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}\\
&=& \lim_{h\to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h} \cdot 
\frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}\\
&=& \lim_{h\to 0} \frac{(x + h + 3) - (x + 3)}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}\\
&=& \lim_{h\to 0} \frac{1}{\sqrt{x + h + 3} + \sqrt{x + 3}}\\
&=& \frac{1}{2\sqrt{x + 3}}.
\end{eqnarray*}
  \item[(c)] (5) Find the equation of the line tangent to
    the graph $y=\ds\sqrt{x + 3}$ at $a=6$.
\item[] {\it Solution:}  By the previous part, the slope of the tangent is $f'(6) = \frac{1}{6}$.
The tangent line has equation:
$$y - 6 = \frac{1}{6}(x - 3)$$
  \end{enumerate}

\end{enumerate}
\item{} Use the sum, product, and/or quotient 
  rules to compute the following derivatives. You may use any correct
  method, but must show work and simplify your answers for full
  credit.

\begin{enumerate}
  \item[(a)] (5) $\ds\frac{d}{dx}\Bigl(\frac{5}{\sqrt{x}} - e^x + 3\Bigr)$
\item[] {\it Solution:}  By the sum, power and exponential rules,
$$\frac{d}{dx}\Bigl(\frac{5}{\sqrt{x}} - e^x + 3\Bigr) = \frac{-5}{2} x^{-3/2} - e^x$$
  \item[(b)] (10) $\ds\frac{d}{du} (u^{5/3} e^u)$
\item[] {\it Solution:}  By the product, power and exponential rules, 
$$\frac{d}{du} (u^{5/3} e^u) = u^{5/3} e^u + \frac{5}{3} u^{2/3} e^u = u^{2/3} \Bigl(u + \frac{5}{3}\Bigr) e^u.$$
  \item[(c)] (10) $\ds\frac{d}{dv}\Bigl(\frac{v^3 - 2v}{v^2 + 5v + 1}\Bigr)$
\item[] {\it Solution:}  By the quotient rule
\begin{eqnarray*}
\frac{d}{dv}\Bigl(\frac{v^3 - 2v}{v^2 + 5v + 1}\Bigr) &=& \frac{(v^2 + 5v + 1)(3v^2 - 2) - (v^3 - 2v)(2v + 5)}{(v^2+5v + 1)^2}\\
&=& \frac{v^4 + 10 v^3 + 5v^2 - 2}{(v^2 + 5v + 1)^2}.
\end{eqnarray*}

  \item[(d)] (5) $\ds\frac{d}{dx}\Bigl(\frac{e^\pi + \pi^e - x^\pi}{4}\Bigr)$
\item[] {\it Solution:}  Since $e^\pi$ and $\pi^e$ are constants, the derivative is just $\ds\frac{-\pi x^{\pi-1}}{4}$.  
\end{enumerate}



\end{enumerate}
\end{document}
