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\begin{document}{\noindent}
\begin{flushright}
Karl Weierstrass \\
MTH 408 HW 6
\end{flushright}

\be

\item[1.]
	Prove that $\ds \lim_{x \to 1} \frac{x^2-x+1}{x+1}=\frac{1}{2}$.
	\newline
	\newline
	We first choose $\varepsilon > 0$.  Then, let $\delta = \text{min} \left( 1,\frac{2}{3}\varepsilon \right)$.
	Suppose that $0 < |x-1| < \delta$.  Then, applying the reverse triangle inequality and that $|x-1| < \delta \leq 1$,
	we see
	\begin{align}
		|x+1|	&= |x-1-(-2)| \notag \\
				&\geq ||x-1|-|-2|| \notag \\
				&= |2-|x-1|| \notag \\
				&> 1. \notag
	\end{align}
	Then, taking recipricols
	\begin{align}
		\left|\frac{1}{x+1}\right| &< 1. \label{E:one} 
	\end{align}
	Further, notice that $|x-1| < \delta \leq 1$ implies that $0 < x < 2$.  Then, it follows easily that
	$-1 < 2x-1 < 3$ and that
	\begin{align}
		|2x-1| < 3. \label{E:two}
	\end{align}
	Our goal is to show that $\left| f(x) - \frac{1}{2} \right| < \varepsilon$.  So, considering $f(x)$ and (\ref{E:one})
	and (\ref{E:two}) above, we see
	\begin{align}
		\left| f(x) - \frac{1}{2} \right|
			&= \left| \frac{x^2-x+1}{x+1} - \frac{1}{2} \right| \notag \\
			&= \left| \frac{2x^2-3x+1}{2(x+1)} \right| \notag \\
			&= \frac{|x-1||2x-1|}{2|x+1|} \notag \\
			&< \frac{|x-1||2x-1|}{2} \notag \\
			&< \frac{\delta \cdot 3}{2} \notag \\
			&< \varepsilon. \notag
	\end{align}
	Thus, $\ds \lim_{x \to 1} \frac{x^2-x+1}{x+1}=\frac{1}{2}$.

\newpage

\item[2.]
	For some $c \in \R$, prove that $\ds \lim_{x \to c} x^3 = c^3$.
	\newline
	\newline
	We will perform a proof by contradiction.  We assume that there exists $c \in \R$ such that
	$\ds \lim_{x \to c} x^3 \neq c^3$.  Then, by the \emph{Sequential Criterion for Limits}, we know that
	there exists a sequence $\{x_n\}$ that converges to $c$ and is not equal to $c$ for all $n$ and that the function
	sequence $\{(x_n)^3\}$ does not converge to $c^3$.  Considering this sequence, which is guaranteed to exist,
	and applying some algebra of limits, we see
	\begin{align}
		\lim_{n \to \infty} x_n^3		&= \left( \lim_{n \to \infty} x_n \right)^3 \notag \\
										&= c^3. \notag
	\end{align}
	This, of course, is a contradiction because we are assuming that the function sequence does not converge to $c^3$.
	Thus, for all $c \in \R$, $\ds \lim_{x \to c} x^3 = c^3$.

\newpage

\item[3.]
	Suppose that $f: \R \to \R$ is a function satisfying $\ds \lim_{x \to 0} f(x) = L$.  For fixed $a > 0$, $g(x)=f(ax)$.
	\be
	\item[a.]
	The graphical relationship between $f$ and $g$ is that the the graph of $f$ is scaled in the horizontal direction
	by $\ds \frac{1}{a}$.  It is important to note that the scaling is ``centered'' at the origin.  That is, the 
	horizontal distance	from a given point to the origin is scaled by this quantity.
	This idea is seen in the next two graphs.
	
	\vspace{.2in}
	\noindent
	Unfortunately, graphics are beyond the scope of this introductory course. Instead, I'll include a matrix and a table
	instead!
	
	\[ A=\begin{bmatrix} 1 & 0 \\ 2 & 4 \end{bmatrix} \mbox{ is a triangular matrix.} \]
	
	
	\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|} 
\hline
& \multicolumn{6}{c|}{Genotype of Parents}\\ \hline
Genotype of Offspring & AA-AA & AA-Aa & AA-aa & Aa-Aa & Aa-aa & aa-aa \\ \hline 
AA &  &  &  & &  &    \\ \hline
Aa &  &  &  & &  &    \\ \hline
aa &  &  &  & &  &    \\ \hline
\end{tabular}
\end{center}
\end{table}
	%\begin{figure}[h]
	%\hfil \includegraphics{c:/images/homework/408hw6_1.eps} \hfil 
	%\end{figure}
	\item[b.]
	We will show that $\ds \lim_{x \to 0} g(x) = L$ is also true.  First assume that $\ds \lim_{x \to 0} f(x) = L$ and
	assume that $a$ is fixed with $a > 0$.  By the \emph{Sequential Criterion for Limits}, we know that if $\{x_n\}$
	is a sequence satisfying that $\{x_n\}$ converges to zero and is not equal to zero
	for all $n$, then $f(x_n)$ converges to $L$.  We will then perform a proof by contradiction.  That is, we assume
	that $\ds \lim_{x \to 0} g(x) \neq L$.  Then, by the \emph{Sequential Criterion for Limits}, we know that there
	must exist a sequence $\{\hat x_n\}$ such that $\{\hat x_n\}$ converges to zero and is not equal to zero
	for all $n$ and $g(\hat x_n)$ does not converge to $L$.  Then, consider the sequence $\{a \hat x_n\}$, which converges
	to zero by algebra of limits and is never equal to zero.  Then, using the sequential argument for $f$ earlier,
	we know that $\{ f(a \hat x_n) \}$ converges to $L$.  But, $\{ f(a \hat x_n) \} = \{ g(\hat x_n)\}$, which we
	are assuming does not converge to $L$.  Thus, we have reached a contradiction and $\ds \lim_{x \to 0} g(x) = L$.
	\ee

\newpage

\item[4.]
	Prove that $\ds \lim_{x \to 0} \sin \left( \frac{1}{x^2} \right)$ DNE
	by using the \emph{Sequential Criterion for Limits}.
	\newline
	\newline
	That is, we must show that two different sequences, $x_n$ and $\hat x_n$, both converge to zero with $x_n \neq 0$ and
	$\hat x_n \neq 0$ for all $n$.  Further, we must show that their function sequences, $f(x_n)$ and $f(\hat x_n)$, 
	converge to $L_1$ and $L_2$ with $L_1 \neq L_2$.
	Consider
	\begin{align}
	\{x_n\} &= \frac{1}{\sqrt{\frac{\pi}{2}+2 \pi n}} \notag \\
	\{f(x_n)\} &= \sin \left( \frac{\pi}{2}+2 \pi n \right) \notag \\
		&= 1 \notag
	\end{align}
	and
	\begin{align}
	\{\hat x_n\} &= \frac{1}{\sqrt{\pi + 2 \pi n}} \notag \\
	\{f(\hat x_n)\} &= \sin \left(\pi + 2 \pi n \right) \notag \\
		&= 0. \notag
	\end{align}
	Clearly, $x_n$ and $\hat x_n$ satisfy the needed properties above.  Further, it is obvious
	that $f(x_n)$ and $f(\hat x_n)$ converge to one and zero respectively.  Thus, we have constructed two different
	sequences such that their function sequences converge to different limits.  And,
	by \emph{Sequential Criterion for Limits}, we have shown that
	$\ds \lim_{x \to 0} \sin \left( \frac{1}{x^2} \right)$ DNE.

\newpage

\item[5.]
	Let $\Q$ denote the set of rational numbers, and define the function
	\[ \chi_{\Q}(x) = 
		\begin{cases}
			1	&\text{if $x \in \Q$}; \\
			0	&\text{if $x \notin \Q$}.
		\end{cases} \]
	Consider $f(x)=x \cdot \chi_{\Q}(x)$. 
	\be
	\item[a.]
	The graph of this function is plotted below.
	%\begin{figure}[h]
	%\hfil \includegraphics{c:/images/homework/408hw6_2.eps} \hfil 
	%\end{figure}
	\item[b.]
	We will next prove that $\ds \lim_{x \to 0} f(x) = 0$.  Choose $\varepsilon > 0$.  Then, let $\delta = \varepsilon$.
	Suppose that $0 < |x-0| < \delta$.  Then, notice that $\chi_{\Q}(x) \leq 1$ for all $x$, by definition.  So,
	\begin{align}
		|f(x) - 0|	&= |x \cdot \chi_{\Q}(x)| \notag \\
					&= |x| \cdot |\chi_{\Q}(x)| \notag \\
					&< \delta \cdot 1 \notag \\
					&= \varepsilon. \notag
	\end{align}
	Thus, we see that $|f(x) - 0| < \varepsilon$.  Then, we have shown that for all $\varepsilon > 0$, there exists
	a $\delta$ such that $0 < |x-0| < \delta$ implies $|f(x) - 0| < \varepsilon$.  And, by definition, we see that
	$\ds \lim_{x \to 0} f(x) = 0$.
	\item[c.]
	For the case when $x_0 \neq 0$, we will show that $\ds \lim_{x \to x_0} f(x)$ DNE.
	To do this, we will consider two different sequences.
	The first, $\{x_n\}$, will be a sequence that converges to $x_0 \neq 0$ along a set of rational numbers and for all
	$n$, $x_n \neq x_0$.
	The second, $\{\hat x_n\}$, will be a sequence that converges to $x_0 \neq 0$ along a set of
	irrational numbers and for all $n$, $\hat x_n \neq x_0$.  It is pretty easy to see that the limits of the function
	sequences are given by
	\begin{align}
		\lim_{n \to \infty}f(\{x_n\}) &= x_0 \notag \\
		\lim_{n \to \infty}f(\{\hat x_n\}) &= 0. \notag 
	\end{align}
	Then, since $x_0 \neq 0$, we have found two sequences in which their function sequences converge to differing values.
	Then, by the \emph{Sequential Criterion for Limits}, the limit of $f(x)$ when $x_0 \neq 0$ does not exist.
	\ee
\ee


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