\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\parindent=0pt
\begin{document}


{\bf Question}

Show that as $x \to \infty$,

\begin{description}

\item[(a)]
$x+o(x)=O(x)$

\item[(b)]
$\{O(x)\}^2=O(x^2)=o(x^3)$

\item[(c)]
$\log(1+o(1))=o(1)$

\end{description}

\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
If $f(x)=O(x)$, then

$\ds \lim_{x \to \infty}\frac{x+f(x)}{x}=1+\lim_{x\to
\infty}\frac{f(x)}{x}=1$

Therefore $x+f(x)=O(x)$

or $x+o(x)=O(x)$.

\item[(b)]
First equality:

Suppose $f(x)$ and $g(x)=O(x),\ \ x \to \infty$.

Then there exists $K_f$ and $K_g>0$ such that

$\left.\begin{array}{cl} & |f(x)| \leq K_f|x|\\  & |g(x)| \leq
K_g|x|\end{array}\right\}$ as $x \to \infty$

Thus $|f(x)||g(x)| \leq K_f K_g|x|^2$

or $|f(x)g(x)| \leq K_h|x^2|$ where $K_h=K_fK_g$

$\begin{array}{rrcl} \Rightarrow & f(x)g(x) & = & O(x^2)\\ \rm{or}
& [O(x^2)]^2 & = & O(x^2) \end{array}$

Second equality:

 Now suppose $\undb{h(x)}_{=f.g}=O(x^2)\ \ x \to \infty$.



Then there exists $K_h$ such that $|h(x)| \leq K_h|x^2|\ \ x \to
\infty$

Thus $\ds0 \leq \lim_{x \to \infty}
\left|\frac{h(x)}{x^3}\right|\leq\lim_{x \to \infty} K_h
\frac{|x^2|}{|x^3|}=K_h \lim_{x \to \infty} \frac{1}{|x|}=0$

Therefore $h(x)=o(x^3)$

or $O(x^2)=o(x^3)$

Hence $[O(x)]^2=O(x^2)=o(x^3)$

\item[(c)]
Suppose $f(x)=o(1)$.

Then $\ds \lim_{x \to \infty} \left|\frac{f(x)}{1}\right|=0$

Thus $\lim_{x\to \infty} \log(1+f(x))=\log 1=0$

$\Rightarrow \un{\log(1+o(1))=o(1)}$
\end{description}



\end{document}