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\begin{document}

\hbox to \hsize{\textbf{Math 331\hfill Homework 3}}
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\centerline{\textit{This homework is due by the end of the day on Friday, September 18}}

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\def\ds{\displaystyle}


\begin{problem}[Textbook problem 1.4.12a]
Suppose that $\lambda$ is the least upper bound of some set $S$, and that
$\lambda$ is \textit{not} in $S$.  Prove that $\lambda$ is an accumulation
point of $S$.  [Hint:  For any $\eps>0$, there is a point $s\in S$ such 
that $\lambda-\eps < s < \lambda$.  Now use the definition of accumulation
point to finish the proof.]
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}
[Textbook problems 1.4.9 and 1.4.10]
\textbf{(a)} Prove lemma 1.4.5: If $x$ is an accumulation point of a set $S$ and if $\eps>0$,
then there is an infinite number of points of $S$ within distance $\eps$ of $S$.
[Hint: Suppose that for some $\eps>0$, there were only a finite number of points,
$s_1,s_2,\dots,s_k$, of $S$ within $\eps$ of $x$, but not equal to $x$.  
Let $\eps'=\min(|s_1-x|,|s_2-x|,\dots,|s_k-x|)$.  Now, show that no $s\in S$
satisfies $0<|s-x|<\eps'$.]
\textbf{(b)} Deduce that if $S$ is a \textbf{finite} subset of $\R$, then
$S$ has no accumulation points.  [This is trivially a corollary of the lemma.]
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}
Prove directly from the epsilon-delta definition of limits, that ${\ds\lim_{x\to 5}}\frac{2x+4}{7}=2$.
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}
Show directly, without using the product rule for limits, that $\ds\lim_{x\to 3}x^3 = 27$.
(Note that $a^3-b^3=(a-b)(a^2+ab+b^2)$.)
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}
Suppose that $\ds\lim_{x\to a}f(x)=L$ and $c\in \R$. Show, using the definition of limit,
that $\ds\lim_{x\to a} cf(x) = cL$.  [Be careful: $c=0$ is a special case.]
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}[Textbook problem 2.2.9]
Suppose that $f(x)\le 0$ for all $x$ in some open interval containing $a$, except possibly 
at $a$.  Suppose that $\ds\lim_{x\to a} f(x) = L$.  Show that $L\le 0$. [Hint: Assume instead
that $L > 0$.  Let $\eps=L/2$ and derive a contradiction.]  (Remark: A similar proof shows that
if $f(x)\ge 0$ for all $x$ near $a$, then $\ds\lim_{x\to a}f(x)\ge0$, if the limit exists.)
\end{problem}

\begin{answer}
\end{answer}



\begin{problem}Let's say that the sum, difference, and constant multiple rules for limits have
already been proved, but that the product rule still needs to be proved.
\begin{enumerate}
\item[(a)] Suppose $\ds\lim_{x\to a}f(x)=L$.  Show (without using the product rule) that
$\ds\lim_{x\to a}f(x)^2=L^2$.
\item[(b)] Use part (a) and the fact that $xy=\frac{1}{4}\big((x+y)^2-(x-y)^2\big)$ to prove
the product rule for limits.
\end{enumerate}
\end{problem}

\begin{answer}
\end{answer}



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