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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions, Discussion 4}
\centerline{October 4, 2000}
\bigskip
\noindent
C 4)  Given:  $z_n = {2 + (-1)^n\over n}$, a sequence converging to 
zero as $n \to \infty$.  Note the terms of this sequence
and their values in $f(x)$ start out like
this:
$$\matrix{n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 &\cdots\cr
&&&&&&&&&&&\cr
z_n & {1\over 1} & {3\over 2} & {1 \over 3} & {3\over 4} & {1\over 5}
& {3\over 6} = {1\over 2} & {1\over 7} & {3\over 8} & {1 \over 9} &{3\over 10}&\cdots\cr
&&&&&&&&&&&\cr
f(z_n) & 1 & {1\over 2} & {1\over 3} & {1\over 2} & {1\over 5} & {1\over 2} 
&{1\over 7} &{1\over 2} & {1\over 3} & {1\over 3} &\cdots \cr}$$
The thing you should notice about this is that there does not seem 
to be any overall pattern to the values $f(z_n)$.  

Indeed, let's consider
the following {\it subsequences}.  Let $n_k$ be the $k$th prime
number bigger than 2.  So $n_1 = 3, n_2 = 5, n_3 = 7, n_4 = 11$, etc.
All these are odd, so $z_{n_k} = {1\over n_k}$.  To compute the value
of the function $f$ on these, we need the second alternative, since
the denominator is the first power of a prime.  $f(z_{n_k}) = z_{n_k}$
and this subsequence of the sequence $f(z_n)$ converges to {\it zero}. 

On the other hand, consider a second subsequence $z_{m_k}$
where $m_k$ is the $k$th even number that is not a power of 2, and is 
not divisible by 3.  So
$m_1 = 10, m_2 = 14, n_3 = 20, n_4 = 22$, and so forth.
For all of the ${m_k}$, we have that $z_{m_k}$ is already in 
lowest terms and equals 3 divided by an even number that is not a power 
of 2.  Hence the last alternative in the definition of $f$ applies
to all of the $z_{m_k}$, and $f(z_{m_k}) = {1\over 3}$ for all $k$.

Since $f(z_n)$ has subsequences converging to different values, it 
cannot be true that $\lim_{n\to\infty} f(z_n)$ exists.  Since $z_n$
is a legal sequence converging to zero (that is, a ``way for $x$
to approach zero'' with $x \ne 0$) this says that $\lim_{x\to 0}f(x)$
cannot exist either.

\bigskip
\noindent
D)  I was looking for something like the following:  In order to use
sequences $x_n \to a$ to decide whether $\lim_{x\to a} f(x) = L$, 
we would have to be able to look at {\it every possible sequence 
$x_n \to a$ (but with $x_n \ne a$ for all $n$)} and decide whether
$\lim_{n\to\infty} f(x_n) = L$.  Let's call those sequences
$x_n\to a$ with $x_n \ne a$ for all $n$  ``legal'' sequences.
In Question A we have three different ``legal'' sequences
converging to $3$, and $f(x_n)$ is converging to $11$ for all 3.
This suggests that $\lim_{x\to 3} f(x) = 11$,
but it is {\it not a complete proof} -- we have not checked that
$\lim_{x\to 3} f(x) = 11$ for $x \to 3$ in all possible ways.  
In Question B, the first two sequences are ``legal'' sequences 
converging to zero, and $\lim_{n\to\infty} f(x_n) = 0$ for each.
This again suggests that $\lim_{x\to 0} f(x) = 0$.  The sequence
from part 4 is {\it not ``legal'' though}, since some of its terms
equal zero (all the odd-numbered terms).  Therefore, the fact
that $f(z_n)$ does not converge {\it does not contradict\/} our expectation
that $\lim_{x\to 0} f(x) = 0$ (!)  Finally, in question C, 
we see that there is a ``legal'' sequence $x_n$ converging
to zero such that $f(x_n)$ converges to 0, another ``legal'' sequence
$y_n$ where $f(y_n)$ converges to ${1\over 7}$, and a third
``legal'' sequence $z_n$ where $f(z_n)$ does not converge at all.
Hence $\lim_{x\to 0}f(x)$ {\it cannot exist\/}.  (The values 
$f(x)$ are not approaching any one limiting value as $x \to 0$.)
\bye


