\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Exam 1}
\centerline{October 11, 2000}
\bigskip
\noindent
I.  Let $A = \{x \in {\bf R} : 1 < x^2 < 9\}$ and let $B$ be the
interval $[-2,4)$.
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\item{A)} What is the set $A\cap B$?
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\item{}{\it Solution:}  $1 < x^2 < 9$ is equivalent to 
either $1 < x < 3$, or $-3 < x < -1$.  So $A$ is the union of two intervals:
$A = (-3,-1) \cup (1,3)$.  The intersection with $B$ is
$$A \cap B = [-2,-1) \cup (1,3)$$
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\item{B)}  Explain why the set $C = \{|x + 1/2| : x \in A\}$
is bounded, and give upper and lower bounds for it.
\bigskip
\item{}  {\it Solution:} Since $C$ is a set of numbers obtained by taking absolute
values, it is bounded below by 0.  Since $x \in A$ implies $|x| < 3$, 
$|x + 1/2| \le |x| + |1/2| < 3 + 1/2$ for all $x \in A$.  Hence
$C$ is also bounded above.
\bigskip
\item{C)}  What are the {\it least upper bound} and {\it greatest
lower bound} for the set $C$ from part B?  Explain your reasoning.
\bigskip
\item{} {\it Solution:}  $|x + 1/2| = |x - (-1/2)|$ is the 
distance from $x$ to $-1/2$ along the number line.  By part A, we know that
$A = (-3,-1) \cup (1,3)$.  Hence the closest points to $-1/2$ in $A$
are $x$ near $-1$, and we can get arbitrarily close to $|x + 1/2| = 1/2$
for $x$ close to $-1$ in $A$.   The greatest lower bound for
$C$ is $1/2$.   Similarly, the farthest points from $-1/2$ are the $x$
near $3$ in $A$.  Hence the least lower bound for
$C$ is $7/2$.
\bigskip
\noindent
II.  For parts A, B see the text or class notes.  {\it Comments:}  There
will be questions like this on the next midterm and final.  {\it You
will be throwing away lots of hard-to-replace points if you do not know 
the definitions, statements of results, and proofs that I put on 
the review sheet for an exam.  Also, part C of this question was
the only problem on this exam intended to be slightly ``tricky''!}
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\item{C)}  Suppose that $x_n$ is a sequence of {\it negative} 
numbers and $x_{n+1}/x_n \le 1$ for all $n \ge 1$.  Show that $x_n$ converges
to some $a \in {\bf R}$.  
\bigskip
\item{} {\it Solution:}  Since all the $x_n$ are negative, when we multiply
both sides of the inequality $x_{n+1}/x_n \le 1$ by $x_n$ to clear 
denominators, the inequality reverses:  $x_{n+1} \ge x_n$ for all $n$.
Also since all of the $x_n$ are negative, the number $M = 0$ is an 
upper bound for the sequence.  By the result of part B, every 
increasing sequence that is bounded above converges to some $a \in {\bf R}$.
The most we can say here is that $a \le 0$.
\vfill\eject
\noindent
III.
\item{A)} Identify  $\lim_{n\to \infty} x_n$ for the sequence 
$$x_n = {3n\over 2n + 5},$$
and prove that $x_n$ converges to your limit using the $\epsilon$ - $n_0$
definition of convergence.
\bigskip
\item{} {\it Solution:}  By our usual methods, the limit should be ${3\over 2}$.
To prove this, note that 
$$\eqalign{\left|{3n\over 2n + 5} - {3\over 2}\right|
&= \left|{-15 \over 4n + 10}\right|\cr
&= {15\over 4n + 10}\cr
&< {15\over 4n}\cr}$$
Hence, given $\epsilon > 0$, if $n \ge n_0 > {15\over 4\epsilon}$, then
${1\over n} < {4\epsilon\over 15}$, so we will have
$$\left|{3n\over 2n + 5} - {3\over 2}\right| < {15\over 4n} < 
{15\over 4}\cdot {4\epsilon\over 15} = \epsilon.$$
By the definition of convergence, this shows that 
$\lim_{n\to\infty} x_n = {3\over 2}$.
\bigskip
\item{B)} Use the ``Big Limit Theorem'' for sequences to 
determine
$$\lim_{n\to\infty} {\sqrt{4n^2 + 3n + 1}\over {\root 3\of {27n^3 + 3}}}$$
\bigskip
\item{} {\it Solution:}  The fastest growing terms on the top and bottom
both are roughly constant times $n$ as $n\to\infty$ (square root 
of $n^2$ and cube root of $n^3$).  Hence, if we divide both top and bottom
by ${1\over n}$, we get
$$\eqalign{
\lim_{n\to\infty} {\sqrt{4n^2 + 3n + 1}\over {\root 3\of {27n^3 + 3}}}
&= \lim_{n\to\infty} {\sqrt{4n^2 + 3n + 1}\cdot{1\over n} \over {\root 3\of {27n^3 + 3}}\cdot{1\over n}}\cr
&= \lim_{n\to\infty} {\sqrt{4 + {3\over n} + {1\over n^2}}\over {\root 3\of {27 + {3\over n^3}}}}\cr
&= {2\over 3}\cr}$$
\bigskip
\item{C)} Consider the sequence
$$x_n = \sin\left({\pi\sqrt{n}\over 2}\right)$$
Find three convergent subsequences of $x_n$ but converging to three different 
limits. (For each one, give a formula for the index sequence $n_k$ for the subsequence).
What does this say about $\lim_{n\to\infty} x_n$?
\bigskip
\item{} {\it Solution:}  The simplest way to find the required subsequences
is to try to find $n_k \to \infty$ where the sine function takes on 
some constant value.  For instance, we know that $\sin(x) = 0$ for
all integer multiplies of $\pi$, so if 
$${\pi\sqrt{n}\over 2} = k\pi$$
for any integer $k\ge 1$, then we get a zero term in the sequence.  From 
the equation above, $n_k = (2k)^2$ gives one such subsequence, with
$x_{n_k} = 0$ for all $k \ge 1$.  Hence $\lim_{k\to\infty} x_{n_k} = 0$.
\item{}
Similarly, to get $\sin(x) = 1$, we can take 
$${\pi\sqrt{n}\over 2} = {(4k+1)\pi\over 2}$$
for all integers $k\ge 1$.  Hence $n_k = (4k+1)^2$ gives a subsequence
converging to $+1$.  Finally, to get $\sin(x) = -1$, we can take 
$${\pi\sqrt{n}\over 2} = {(4k+3)\pi\over 2}$$
for all integers $k\ge 1$.  Hence $n_k = (4k+3)^2$ gives a subsequence
converging to $-1$.
\item{}
The fact that we can find these subsequences says that the whole
$x_n$ sequence {\it does not converge}.
\bye