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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 2 -- April 5, 2013}
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\noindent
I.  
\item{A)}  (15)  Show using the $\varepsilon, n_0$ definition that 
$$\lim_{n\to\infty} {5n + 1\over 3n + 3} = {5\over 3}.$$  
\item{} {\it Solution:}  Given $\varepsilon > 0$, let $n_0 > {4\over 3\varepsilon}$.  Then
for all $n \ge n_0$,
$$\left| {5n + 1\over 3n + 3} - {5\over 3}\right| = {4\over 3n + 3} < {4\over 3n} 
\le {4\over 3n_0} < \varepsilon.$$
This shows the limit is ${5\over 3}$.  
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\item{B)}  (15)  Show using the $\varepsilon,\delta$ definition that
$$\lim_{x\to 1} {5x + 1 \over 3x + 3} = 1.$$  
\item{} {\it Solution:}  Given $\varepsilon > 0$, let 
$\delta = \min\left(1,{3\varepsilon\over 2}\right)$.
Then for all $x$ in the deleted neighborhood of $1$ defined by $0 < |x - 1| < \delta$
we have $0 < x < 2$, so ${1\over |x+1|} < 1$ and $|x - 1| < {3\varepsilon\over 2}$. 
Therefore  
$$\left| {5x + 1\over 3x + 3} - 1\right| = {2 |x - 1| \over 3 |x + 1|} =
{1\over |x + 1|} \cdot {2\over 3} \cdot |x - 1| < 1\cdot {2\over 3}\cdot {3\varepsilon\over 2} = 
< \varepsilon .$$
This shows the limit is $1$.  
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\noindent
II.  
\item{A)} (20) State and prove the Monotone Convergence Theorem
for sequences.  (You may give the proof in the case that the 
sequence is monotone increasing.)
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\item{} {\it Solution:}  The statement is that any monotone bounded
sequence of real numbers converges.  If the sequence is monotone 
increasing, let $a = {\rm lub}\{x_n \mid n \in {\bf N}\}$.  Then 
for all $\varepsilon > 0$, $a - \varepsilon$ is not an upper bound 
for $\{x_n\mid n \in {\bf N}\}$, so there exist $n_0$ such that $a - \varepsilon < x_{n_0} \le a$.
But then since $\{x_n\}$ is monotone increasing, we have 
$a - \varepsilon < x_{n_0} \le x_n \le a$ for all $n\ge n_0$.  This shows
$x_n \to a$, since it implies $|x_n - a| < \varepsilon$ for all $n\ge n_0$.  
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\item{B)} (10)  Let $\{x_n\}$ be the sequence defined by $x_1 = 1$ and 
$$x_{n+1} = {5 x_n + 3\over 16}$$
for all $n \ge 1$.  Does this sequence converge?  Why?
If it does, what is the limit?
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\item{} {\it Solution:}  We have $x_1 = 1$ and $x_2 = {5\cdot 1 + 3\over 16} = {1\over 2}$.  
Assuming $x_{k+1} < x_k$, it follows that 
$$x_{k+2} = {5x_{k+1} +3 \over 16} < {5x_k + 3\over 16} = x_{k+1}.$$
Therefore, the sequence is monotone (strictly) decreasing.  The terms are clearly
bounded below by $0$, so the sequence converges by the Monotone Convergence
Theorem.  The limit is found by letting $n \to \infty$ in the recurrence.  If
$a$ denotes the limit then $a = {5a + 3\over 16}$, so $a = {3\over 11}$.  
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\noindent
III. (10)  Let $x_n = \sin(n) - \cos(n)$ (the sequence of values
of $\sin(x) - \cos(x)$ at the angles $x$ given by a whole number 
$n\ge 1$ in radian measure).  Show that there exists a subsequence 
$x_{n_k} = \sin(n_k) - \cos(n_k)$ that converges.  State any ``big theorems''
that you are using.
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\noindent
{\it Solution 1:}  We have 
$$|x_n| = |\sin(n) - \cos(n)| \le |\sin(n)| + |\cos(n)| \le 2$$ 
for all $n$ by the triangle inequality.  This shows the sequence is bounded.
Hence the Bolzano-Weierstrass theorem implies it has a convergent subsequence.
(Note:  This is not the tightest possible upper bound.  In fact, by trig identities,
$$\sin(n) - \cos(n) = \sqrt{2}\sin\left(n - {\pi\over 4}\right)$$
so $|\sin(n) - \cos(n)| \le \sqrt{2}$ is also true for all $n$.  There are $n$ for
which $|\sin(n) - \cos(n)| > 1$, though, so that is not a valid upper bound.)
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\noindent
It is also possible to argue like this:
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\noindent
{\it Solution 2:}  Since $|\sin(n)| \le 1$ for all $n$, by the Bolzano-Weierstrass
theorem there exists an index sequence $n_k$ such that $\sin(n_k)$ converges to some
$a \in {\bf R}$.  
There is no guarantee that $\cos(n_k)$ also converges, but we can apply the 
Bolzano-Weierstrass theorem again to that sequence and get a subsequence 
of the index sequence $n_k$, say $n_{k_\ell}$ for some strictly increasing
sequence $k_\ell$ of $k$-values, such that $\cos(n_{k_\ell})$ converges to
some $b \in {\bf R}$.  
But then $\sin(n_{k_\ell})$ is a subsequence of a convergent sequence, so 
it also converges to $a$.  Then by the limit sum theorem, as $\ell\to\infty$, 
$$\sin(n_{k_\ell}) - \cos(n_{k_\ell}) \to a - b.$$
(Note: Solution 1 is certainly more direct(!))
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\noindent
IV.  Give an example, or give a reason why there can be no
such examples:
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\item{A)} (10)  A function $f$ that is continuous at all real $c$, 
with $f(x) = 3$ for all $x \in {\bf Q}$, but $f(\sqrt{2}) = 4$.  
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\item{} {\it Solution:}  There are no such examples because we can 
find sequences $x_n \to \sqrt{2}$ where all the $x_n$ are rational.
If $f$ is continuous at $c = \sqrt{2}$, then $f(x_n) \to f(\sqrt{2}) = 4$ but $f(x_n) = 3$
for all $n$ if $x_n$ is rational, so this is impossible.
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\item{B)} (10)  A function $f$ that is continuous and bounded on an open 
interval $(a,b)$ with $M = {\rm lub} \{f(x) \mid x \in (a,b)\}$, but such that
$f(x) \ne M$ for all $x \in (a,b)$.  
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\item{} {\it Solution:}  An example is $f(x) = x$ on $(0,1)$.  We have
$M = 1$, but $f(x) \ne 1$ for any $x \in (0,1)$.  (The Extreme Value Theorem
does not apply on open intervals.)  
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\item{C)} (10)  A continuous function on $[0,1]$ with $f(0) = 4$, $f(1) = -2$, 
but $f(x) \ne 0$ for all $x \in (0,1)$.  
\item{} {\it Solution:}  There are no such examples because the Intermediate
Value Theorem says that there must be solutions of all equations $f(c) = k$
for $k$ between $f(0) =4$ and $f(1) = -2$, including $k = 0$.   
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\noindent
{\it Extra Credit.} (10) A function is said to be {\it right-continuous}
at $c$ if the one-sided limit $\lim_{x \to c^+} f(x)$ exists and equals $f(c)$.   
True or False (and give a proof or
a counterexample):  If $f$ is
right-continuous at all $c \in [a,b]$ for some closed interval $[a,b]$, then
$\{f(x) \mid x\in [a,b]\}$ is a bounded subset of ${\bf R}$.   
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\noindent
{\it Solution:}  This is not true.  An example is 
$$f(x) = \cases{0 & if $x \ge 0$\cr
                1/x & if $x < 0$\cr}$$
for $x \in [-1,1]$.
Then $\lim_{x \to c^+} f(x) = f(c)$ for all $c \in [-1,1]$, but the set of 
values is not bounded, since $\lim_{x\to 0^-} f(x) = -\infty$.  
\bye
