

\begin{center} 
{\large \bf PROBLEM SET 19.1}
\end{center}
\vspace{.2in}

\noindent
{\bf For each of the polynomials in Problems 1--3, write
the Taylor polynomial  of the same degree
at the given value of $ \boldmath \mbox{$a$} $
and show algebraically that the two polynomials
are identical.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $
$ p (x) = 3x + 5 , ~~ a = 1 $ 
& \hspace{1in} &
$ ~ \: ${\bf 2.} $ ~~ $ 
$ p (x) = x^5 - 1 , ~~ a = 1 $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ p (x) = ( 2x - 3 )^3 , ~~ a = - 2 $ &&
\end{tabular} 
\vspace{.2in}

\noindent 
$ ~~ \: ${\bf 4.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Write the third degree Taylor polynomial for $ f(x) = 
1 / \sqrt{ x } $ at $ x = 81 $
and use it to approximate $ \sqrt{ 83 } $.
Compared with the calculator value of $ \sqrt{ 83 } $;
what is the error?
\end{minipage}
\vspace{.2in}

\noindent 
$ ~~ \: ${\bf 5.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Write the fifth degree Taylor polynomial for $ f(x) = 
\ln \: ( 1 + x ) $ at 
$ x = 1 $.  Use it to approximate $ 1 / \sqrt{ 83 } $.
What is the error?
\end{minipage}
\vspace{.2in}

\noindent 
{\bf In Problems 6--9, calculate the Taylor
polynomial of the indicated degree at
$ \boldmath \mbox{$a$} $}.
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
$ ~ \: ${\bf 6.} $ ~~ $
$ f (x) = x \: \ln \: x ; ~~ n = 3 , ~~ a = 1 $
& \hspace{1in} & 
$ ~ \: ${\bf 7.} $ ~~ $
$ h (t) = \sin \: t ; ~~ n = 4 , ~~ a = \frac{ \pi }{ 3 } $
\\ [0.2in]
$ ~ \: ${\bf 8.} $ ~~ $
$ f(t) = \sqrt{ \frac{ 1 + t }{ 1 - t }} ; ~~ n = 2 , ~~ a = 0 $ 
&& $ ~ \: ${\bf 9.} $ ~~ $
$ f (x) = \sin ^2 \: x ; ~~ n = 5 , ~~ a = \pi $
\end{tabular}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 10.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
If $ p (x) = 3 x^3 + 7x^2 
+ 10 x + 2 $ is the third degree Taylor polynomial 
at $ a = - 1 $ for $ f $, find $ f^{''} ( - 1 ) $.
\end{minipage}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 11.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Show that $ e^x > 1 + 
x + {\displaystyle \frac{ x^2 }{ 2 ! } + 
\frac{ x^3 }{ 3 ! }} $ if $ x > 0 $.
[Hint:  Show that the function $F$, obtained
by subtracting the polynomial from $ e^x $,
and its first three derivatives are zero at $ x = 0 $
and that the fourth derivative is positive.
Then integrate.]  
\end{minipage}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 12.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
On a single coordinate system plot 
$ \sin \: x $ and its Taylor polynomials
$ P_1 , P_3 $, and $ P_5 $ at $ a=0 $.
\end{minipage}
\vspace{.2in}

\newpage
 

\begin{center} 
{\large \bf PROBLEM SET 19.2}
\end{center}
\vspace{.2in}

\noindent
{\bf For each of the functions in Problems 1--5, write
Taylor's formula with remainder for the given values of
$ \boldmath \mbox{$a$} $ and $ \boldmath \mbox{$n$} $.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $
$ {\displaystyle \frac{ 1 }{ 1 - x }} ; ~~ a = 0 , ~~
n = 6 $ & \hspace{1in} &
$ ~ \: ${\bf 2.}
$ ~~ $ $ ( + x )^{ - 1/2 } ; ~~
a = 0 , ~~ n = 4 $ \\
\\ [0.2in]
$ ~ \: ${\bf 3.}
$ ~~ $ $ \ln \: {\displaystyle \frac{ 1 + x }{ 1 - x }} ; ~~
a = 0 , ~~ n = 5 $ &&
$ ~ \: ${\bf 4.}
$ ~~ $ $ \sin \: x ; ~~ a = 0 , ~~ n = 6 $ 
\\ [0.2in]
$ ~ \:${\bf 5.} $ ~~ $ $ \cos \: x 3x ; ~~
a = \frac{ \pi }{ 2 } , ~~ n = 4 $ && 
\end{tabular} 
\vspace{.2in}

\noindent 
{\bf In Problems 6--8, use 
Taylor's formula with remainder at 
$ \boldmath \mbox{$ a = 0 $} $
with 
$ \boldmath \mbox{$ n = 3 $} $
to approximate the indicated quantity.  Estimate
the error and round your approximation to the number of decimal 
places justified.}
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
$ ~ \:${\bf 6.} $ ~~ $
$ \sqrt{ 1.13 } $ & \hspace{2.2in} & 
$ ~ \:${\bf 7.} $ ~~ $ $ e^{ 0.2 } $ 
\\ [0.2in]
$ ~ \:${\bf 8.} $ ~~ $ $ \ln \: 0.98 $
&&
\end{tabular}
\vspace{.2in}

\noindent 
{\bf In Problems 9--11, use 
Taylor's formula with remainder 
to find the best (rounded) four decimal place approximation 
to the given quantity.  As a first step, choose
$ \boldmath \mbox{$ n $} $
so that $ \boldmath \mbox{$ | R_n (x) | < 
0.5 \times 10^{ - 4 } $} $.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll} 
$ ~ \:${\bf 9.} $ ~~ $ $ \sqrt{ 1.2 } $ 
& \hspace{2.2in} & {\bf 10.} $ ~~ $
$ \cos \: 29 ^{ \circ } $
\\ [0.2in]
{\bf 11.} $ ~~ $ $ \ln \: 1.25 $ && 
\end{tabular}
\vspace{.2in}

\noindent 
$ ~ \:${\bf 12.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
The problem is to find the best four decimal place 
approximation to
\[ {\displaystyle \int_{ 0 }^{ 0.3 }} \: 
\sin ^4 \: \theta \: d \theta . \]
Compare the effort in (a) using the approximation 
$ \sin \: \theta \approx \theta $,
(b) approximating with Simpson's method based on two
parabolas, and (c) evaluating the integral analytically.
\end{minipage}
\vspace{.2in}

\noindent 
$ ~ \:${\bf 13.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Approximate the integral $ {\displaystyle \int_0^{ 0 \cdot 4 }} \:
x \sqrt{ \cos \: x } \: d x $ in two ways,
first by using $ \cos \: x \approx 1 
- \frac{ x^2 }{ 2 } $ in the integrand and then with
Simpson's method based on four parabolas.
\end{minipage}
\vspace{.2in}

 
\newpage

 
\begin{center} 
{\large \bf PROBLEM SET 19.3}
\end{center}
\vspace{.2in}

\noindent
{\bf For Problems 1-9, write
L'Hopital's Rule to find the indicated limits.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $
$ {\displaystyle \lim_{ x \rightarrow 2 }} \:  
{\displaystyle \frac{ x^2 + x - 6 }{ 2x^3 
- 3x^2 - 4x + 4 }} $
& \hspace{1in} &
$ ~ \: ${\bf 2.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow 0 }} \:
{\displaystyle \frac{ e^x - e^{ - x } }{ \sin \: x }} $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \pi }} \:
{\displaystyle \frac{ \sin \: m x }{ \sin \: n x }} $; 
$m,n$ integers
&& $ ~ \: ${\bf 4.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow 0 }} \:  
{\displaystyle \frac{ 1 - \cos \: 5x }{ x }} $
\\ [0.2in]
$ ~ \: ${\bf 5.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow 0 }} \:
{\displaystyle \frac{ e^x - e^{ - x } - 2x }{ x 
- \sin \: x }} $
&& 
$ ~ \: ${\bf 6.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:
{\displaystyle \frac{ \frac{ \pi }{ 2 } - 
\mbox{Arctan} \: x }{ \ln \: 
\left ( 1 + \frac{1}{x} \right ) }} $
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $ 
$ {\displaystyle \lim_{ \alpha \rightarrow \pi }} \: 
{\displaystyle \frac{ 1 + \tan \: 
\frac{ 3 \alpha }{ 4 } }{ \mbox{cot} \: 
\frac{ \alpha }{ 2 } }} $
&& 
$ ~ \: ${\bf 8.} $ ~~ $ 
$ {\displaystyle \lim_{ \alpha \rightarrow \frac{ \pi }{ 2 } }} \:  
{\displaystyle \frac{ \sin ^2 \: 2 \alpha }{ 1 
+ \cos \: 2 \alpha }} $
\\ [0.2in]
$ ~ \: ${\bf 9.} $ ~~ $ 
$ {\displaystyle \lim_{ \theta \rightarrow 0 }} \: 
{\displaystyle \frac{ \theta - \sin \: ( \sin \: \theta ) }{ \theta 
- \sin \: \theta }} $
&&
\end{tabular} 
\vspace{.2in}

\noindent 
{\bf In Problems 10 and 11, formulate and solve a limit 
problem that, to your mind, substantiates the claim.}
\vspace{.2in}

\noindent 
{\bf 10.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
$ \mbox{Arctan} \: \approx x $
for small values of $x$.
\end{minipage}
\vspace{.2in}

\noindent 
{\bf 11.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
The area of the region bounded by the tangent curve and 
the sine curve from 0 to a small positive number $x$ is about
three times the corresponding area of the region between 
the line $ y = x $ and the sine curve.
\end{minipage}
\vspace{.2in}

\noindent 
{\bf 12.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Show that $ {\displaystyle \lim_{ x \rightarrow 0 }} \:
{\displaystyle \frac{ {\displaystyle \int_0^x } \:
\frac{ \sin \: t }{ t } \: d t }{ x }} = 1 $. 
\end{minipage}
\vspace{.2in}

\newpage 
 
\begin{center} 
{\large \bf PROBLEM SET 19.4}
\end{center}
\vspace{.2in}

\noindent
{\bf For Problems 1-10, find the limits.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  
{\displaystyle \frac{ x^4 }{ e^x }} $
& \hspace{1in} &
$ ~ \: ${\bf 2.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  
{\displaystyle \frac{ \ln \: ( 1 + e^{ - 3x } ) }{ x }} $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  x^3 \left (
\frac{ \pi }{ 2 } - \mbox{Arctan} \: x - 
\frac{ 1 }{ x } \right ) $
&&
$ ~ \: ${\bf 4.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow 0^+ }} \:  
( 1 + x )^{ \ln \: x } $
\\ [0.2in]
$ ~ \: ${\bf 5.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  x^3 
\left ( \frac{ 1 }{ 2 } \: \ln \: 
\frac{ x + 1 }{ x - 1 } - 
- \frac{ 1 }{ x } \right ) $
&& 
$ ~ \: ${\bf 6.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  \left ( 1 
+ \frac{ t }{ x } \right )^x $
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  \left ( 
{\displaystyle \frac{ 1 + \frac{ 1 }{ 2x } }{ 1 - 
\frac{ 1 }{ 2x } }} \right )^x $
&& $ ~ \: ${\bf 8.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  \left ( 
\frac{ 1 }{ x - 1 } 
- \frac{ 1 }{ \ln \: x } \right ) $
\\ [0.2in]
$ ~ \: ${\bf 9.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow \infty }} \:  
\left ( \frac{ x + 4 }{ x - 4 } \right )^x $
&& 
{\bf 10.} $ ~~ $ 
$ {\displaystyle \lim_{ x \rightarrow 0 }} \: 
\left [ \frac{ 1 }{ \sin ^2 x } - 
\frac{ 1 }{ x^2 } \right ] $
\end{tabular} 
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 11.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Translate each of the following statements into a limit 
problem involving ratios.  Solve the problems.

\begin{enumerate}
\begin{enumerate}
\item As $ x \rightarrow \infty , ~~ \ln \: x $ increases at a 
much slower rate than any positive power of $x$.

\item
As $ x \rightarrow \infty , ~~ e^x $ increases at a much faster rate than
any positive power of $x$.

\end{enumerate}
\end{enumerate}
\end{minipage}
 

