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\begin{document}

\hbox to \hsize{\textbf{Math 331\hfill Homework 4}}
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\centerline{\textit{This homework is due by the end of the day on Saturday, September 26}}

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\def\ds{\displaystyle}


\begin{problem}
Suppose that $f(x)$ is defined and bounded on an open interval containing $0$, except possibly at $0$ itself.
(That is, there is a number $B$ such that $|f(x)|<B$ for all $x$ in that interval, except possibly $x=0$.)
Show that $\ds\lim_{x\to 0}xf(x)=0$.  [Hint: Use the Squeeze Theorem and the fact that $|x|$ is a continuous
function.]
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}



\begin{problem}
Suppose that $f(x)$ is a continuous function.  Show that $|f(x)|$ is also continuous.
[Hint: This is trivial.  Apply known facts.]
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}



\begin{problem}[Textbook problem 2.5.7]
Suppose that $f$ is continuous at $a$ and that $f(a)>0$.  Show that there is a $\delta>0$ such
that $f(x)>0$ for all $x$ in the interval $(a-\delta,a+\delta)$.  [Hint: Let $\eps=f(a)$.]
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}



\begin{problem}[Textbook problem 2.6.7b]
Show that $p(x)=x^4-x^3+x^2+x-1$ has at least two roots in the interval $[-1,1]$.
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}


\begin{problem}
Prove the second half of the Extreme Value Theorem.  That is, show that if $f$ is continuous
on a closed, bounded interval $[a,b]$, then there is a $x_o\in[a,b]$ such that $f(x)\ge f(x_o)$
for all $x\in[a,b]$.  [Hint: An easy way to do this is to consider the function $g(x)=-f(x)$.]
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}


\begin{problem}
Show that any linear function $f(x) = mx+b$ is uniformly continuous on $\R$.
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}


\begin{problem}
Let $f(x)=\frac{1}{x}$.
\begin{itemize}
\item[(a)] Show that  for any $c>0$, $f(x)$ is uniformly continuous on $[c,\infty)$,
\item[(b)] Show that $f(x)$ is not uniformly continuous on $(0,\infty)$.
\end{itemize}
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}



\begin{problem}[Textbook problem 2.6.12ab]
We say that a function $f$ satisfies a \textbf{Lipschitz condition} if there is a positive
real number $M$ such that for all $x,y\in\R$, $|f(x)-f(y)|<M|x-y|$.
\begin{enumerate}
\item[(a)] Show that if $f$ satisfies a Lipschitz condition, then $f$ is uniformly 
     continuous on $(-\infty,\infty)$. 
\item[(b)] Show that $f(x) = (1+|x|)^{-1}$ is uniformly continuous on $(-\infty,\infty)$
     by showing that it satisfies a Lipschitz condition.
     [Hint: Recall that $\big||x|-|y|\big|\le|x-y|$.]
\end{enumerate}
\end{problem}

\begin{answer}
%You can put your answer here
\end{answer}






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