\magnification=1200
\baselineskip=12pt
\nopagenumbers

\noindent
MATH 180\hfill FALL 1996

\noindent


\bigskip\noindent
\centerline{ASSIGNMENT \#19}

\bigskip\noindent
3.4, \#10.
Change in income $=\int^{12}_0 r(t)dt
=\int^{12}_0 40(1.002)^t dt = \$485.80$

\bigskip\noindent
3.4, \#14.
Notice that each square on the graph represents 10/6, or 5/3
miles.  At t=1/3 hours, $v=0$.  The area between the $v$ graph
and the $t$-axis over the interval $0<t<1/3$ is
$-\int_0^{1/3}v dt$ which is about one square or 5/3 miles.
At $t=1/3$ she is about $5-5/3$ or about 10/3 miles from the lake.
$v$ is positive on the interval $1/3 \leq t \leq 1$, so she
is moving away from the lake on that interval. About 8 squares
are between the $v$ graph and the $t-$axis.  At $t=1$,


$$\displaystyle \int^1_0 v\; dt= \int^{1/3}_0 v\; dt+
\int_{1/3}^1 v\; dt\approx -{5\over 3}
+8\cdot {10\over 6}={35\over 3},$$

and the cyclist is about $5+{35\over 3}={50\over 3}
=16{2\over 3}$ miles from the lake.  Since, starting from the
moment $t={1\over 3}$, she moves away from the lake, the
cyclist will be farthest away from the lake at $t=1$.
The maximal distance is $16{2\over 3}$ miles.


\vskip 2.5in


\bigskip\noindent
3.5, \#2.
$$\lim_{t\to\infty} t^3e^{-t}=0$$
A viewing window with $0\le x\le 20$
(or larger $x$) and $0\le y\le 2$ suggests these limits.
(In fact, $t^ne^{-1}\rightarrow 0$ as $t\rightarrow\infty$ for any
$n>0$.  This expresses in another way the fact
that exponential functions dominate power functions.)


\eject

\bigskip\noindent
3.5, \#3.
$$\lim_{x\to \infty} {\sin x\over x}=0$$
A viewing window with $0\le x\le 20$
(or larger $x$) and $-2\le y\le 2$ suggests these limits.
It is easy to give a direct argument why the limit is 0.
The numerator $\sin x$ is always between $-1$ and $1$ while the
denominator $x$ gets arbitrarily large as $x\rightarrow\infty$.

\bigskip\noindent
3.5, \#5.
If the exponent were a fixed number, like 3, we could bring
the limit inside the expression that is raised to the exponent. However,
the exponent $n$ is not fixed, but rather is going to infinity
at the same time that the 1+${1\over n}$ is going to one.
So we cannot bring the limit inside the exponent.

Now, if we were to ask the calculator to evaluate
$(1+{1\over n})^n$ for $n=$ say $10^{20}$, it will try to evaluate
$1+{1\over 10^{20}}$ and it will have no choice but to round off the
answer to exactly 1.  Then it will take $1^{10^{20}}$, which is 1.
\end