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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 2 -- March 28, 2014}
\bigskip
\noindent
I.  
\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.)
\medskip
\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 
implies $|x_n - a| < \varepsilon$ for all $n\ge n_0$.  Hence 
$x_n \to a$.  
\bigskip
\item{B)} (10)  Let $\{x_n\}$ be the sequence defined by $x_1 = 1$ and 
$$x_{n+1} = \sqrt{3x_n + 1}$$
for all $n \ge 1$.  Does this sequence converge?  Why?
If it does, what is the limit?
\bigskip
\item{} {\it Solution:}  We have $x_1 = 1$ and $x_2 = \sqrt{3\cdot 1 + 1} = 2$.  
Assuming $x_{k+1} > x_k$, it follows that 
$$x_{k+2} = \sqrt{3x_{k+1} + 1} > \sqrt{3x_k + 1} = x_{k+1}.$$
Therefore, the sequence is monotone (strictly) increasing.  Next, we claim that all the 
terms in the sequence are bounded above by $4$.  This is true for $n = 1,2$ by the above.
Furthermore, if $x_k < 4$, then $x_{k+1} < \sqrt{3\cdot 4 + 1} = \sqrt{13} < 4$ so the sequence is bounded above by $4$.  Therefore, it converges by the Monotone Convergence
Theorem.  The limit is found by letting $n \to \infty$ in the recurrence.  If
$a$ denotes the limit then $a = \sqrt{3a + 1}$, so $a^2 - 3a - 1 = 0$.  By the 
quadratic formula, $a = {3 + \sqrt{13}\over 2}$ (the negative sign gives a number $< 0$, so
that cannot be the limit since $x_n > 0$ for all $n$).    
\bigskip
\noindent
II.  
\item{A)}  (15)  Show using the $\varepsilon, n_0$ definition that 
$$\lim_{n\to\infty} {4n + 1\over 7n + 3} = {4\over 7}.$$  
\item{} {\it Solution:}  Given $\varepsilon > 0$, let $n_0 > {5\over 49\varepsilon}$ (such
$n_0$ exist since ${\bf N}$ is not bounded in ${\bf R}$).  This implies that 
${5\over 49n_0} < \varepsilon$.  Then for all $n \ge n_0$,
$$\left| {4n + 1\over 7n + 3} - {4\over 7}\right| = {5\over 49n + 21} < {5\over 49 n} 
\le {5\over 49 n_0} < \varepsilon.$$
This shows the limit is ${4\over 7}$.  
\bigskip
\item{B)}  (15)  Show using the $\varepsilon,\delta$ definition that
$$\lim_{x\to 1} {4x + 1 \over 7x + 3} = {1\over 2}.$$  
\item{} {\it Solution:}  Given $\varepsilon > 0$, let 
$\delta = \min(1,6\varepsilon)$.
Then for all $x$ in the deleted neighborhood of $1$ defined by $0 < |x - 1| < \delta$
we have $0 < x < 2$, so ${1\over |14x+6|} < {1\over 6}$ and $|x - 1| < {6\varepsilon}$. 
Therefore  
$$\left| {4x + 1\over 7x + 3} - {1\over 2}\right| = {|x - 1| \over |14x + 6|} =
 < 6\varepsilon \cdot {1\over 6} = \varepsilon .$$
This shows the limit is ${1\over 2}$.  
\bigskip
\noindent
III. (10) Suppose $\{x_n\}$ is a sequence such that 
$|x_n - 20| < 50$ for all $n \ge 1$.  Show that there exists
some number $a \in [-30,70]$ and a subsequence 
$\{x_{n_k}\}$ such that $x_{n_k} \to a$.  State any ``big theorems''
you are using.
\bigskip
\noindent
{\it Solution:}  We have $-30 < x_n < 70$ for all $n$.
This shows the sequence is bounded.
Hence the Bolzano-Weierstrass theorem implies it has a convergent subsequence
$x_{n_k} \to a$.  By the Order Limit theorem, it follows that $-30 \le a \le 70$.  

\bigskip
\noindent
IV.  Give an example, or give a reason why there can be no
such examples:
\bigskip
\item{A)} (10)   A function $f$ that is continuous at $x = 0$, but
not continuous at any other $x$.   
\bigskip
\item{} {\it Solution:}  The function 
$$f(x) = \cases{ x & if $x \in {\bf Q}$\cr -x & if $x\notin {\bf Q}$\cr}$$
is an example.  $\lim_{x \to 0} f(x) = 0 = f(0)$, so $f$ is continuous
at $x =0$.  However, $f$ is not continuous at any other $c$, since 
all neighborhoods of any $c \ne 0$ contain $x$ where $f(x)$ is close
to $c$ and other $x$ where $f(x)$ is close to $-c$.  Therefore $\lim_{x \to c} f(x)$
does not exist if $c \ne 0$.  
\bigskip
\item{B)} (10)   A sequence ${x_n}$ such that $x_n \to 3$, but 
for all $n_0 \in {\bf N}$ there exist $n \ge n_0$ with $x_n < 0$.  
\bigskip
\item{} {\it Solution:}  There cannot be such a sequence.  Let $\varepsilon = 1$,
for instance.  Then if $x_n \to 3$, there must equal an $n_0 \in {\bf N}$ such
that $|x_n - 3| < 1$ for all $n \ge n_0$.  But $|x_n - 3| < 1$ implies $2 < x_n < 4$
so all the terms for $n \ge n_0$ must be strictly positive.  
\bigskip
\item{C)} (10) A function $f$ and a $c$ in the domain of $f$
such that $\lim_{x\to c^+} f(x) = \lim_{x\to c^-}f(x)$, but 
$f$ is not continuous at $c$.   
\item{} {\it Solution:}  Let 
$$f(x) = \cases{1 & if $x \ne 0$\cr
                0 & if $x = 0$\cr}$$
Then for $c = 0$, we have  $\lim_{x\to 0^+} f(x) = 1 = \lim_{x\to 0^-}f(x)$.
But $f(0) = 0 \ne 1$.  Therefore $f$ is not continuous at $0$.  (This is 
a {\it removable discontinuity}.)
\bigskip
\noindent
{\it Extra Credit.} (10) Assume that $\{x_n\}$ is a sequence that 
converges to $a$.  Construct a new sequence $\{y_n\}$
by making $y_n$ the average of the first $n$ terms in the original sequence:
$y_n = {x_1 + x_2 + \cdots + x_n\over n}.$
Show that $\{y_n\}$ also converges to $a$.    
\bigskip
\noindent
{\it Solution:}  Intuitively, the idea is that as $n$ gets big, most of the 
terms in the numerator of $y_n$ will be getting close to $a$, but a few 
terms at the start might not be that close.  However, the $n$ in the denominator
means that those few terms at the start contribute less and less as $n \to \infty$.   
But we need a more precise way to say this to get an actual proof.  Here's an idea.  
By the triangle inequality, note that 
$$\eqalign{
|y_n - a| &= \left|{x_1 + x_2 + \cdots + x_n\over n} - a\right|\cr
 &= {1\over n} | x_1 + x_2 + \cdots + x_n - na|\cr
&\le {1\over n} \left(|x_1 - a| + \cdots + |x_n - a|\right).\cr}$$
Since $\{x_n\}$ converges to $a$, recall that we know the sequence is
bounded.  This implies that there exists $M$ such that
$|x_n - a| \le M$ for all $n$ as well.  Now, given $\varepsilon > 0$, 
there exists $N$ such that $|x_n - a| < \varepsilon/2$ for all $n \ge N$.
So if $n \ge N$, continuing from the last line above,
$$\eqalign{
&\le {1\over n}(|x_1 - a| + \cdots + |x_{N-1} - a|) + {n - N + 1\over n} {\varepsilon\over 2}\cr
&< {N M\over n} + {\varepsilon\over 2}.\cr}$$
Since we can think of the $N$ that works here as fixed, but $n$ is still 
allowed to grow, note that we can now find $n$ large enough
so that ${N M\over n} < {\varepsilon\over 2}$ as well by taking $n > {2NM\over \varepsilon}$.
In other words, given $\varepsilon > 0$, we take 
$n_0 = \max(N,{2NM\over \varepsilon})$.  Then $n \ge n_0$ implies $|y_n - a| < \varepsilon$.  
This shows that $y_n \to a$ as $n\to \infty$.   
\bye
