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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}


{\bf Question}

For $|z| \to \infty$ find the sectors in the complex plane where
the following order estimates are satisfied for all positive $n$.

\begin{description}
\item[(a)]
$z^n=o(e^z)$

\item[(b)]
$e^z=o(z^n)$

\item[(c)]
$z^n=o(e^{z^2})$

\end{description}


\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds z^n=o(e^z) \ {\rm as} \ |z|\to\infty \Rightarrow
\lim_{|z|\to\infty}\left|\frac{z^n}{e^z}\right|=0,$ for all $n>0$.

This is true for $Re(z)>0,\ (|e^z| \gg 1)$

$\Rightarrow z^n=o(e^z),\ |z|\to\infty,\ Re(z)>0$

\item[(b)]
$e^z=o(z^n) \Rightarrow
\lim_{|z|\to\infty}\left|\ds\frac{e^z}{z^n}\right|=0$ for all
$n>0$

When $Re(z)<0,\ \ |e^z| \ll 1$.

Therefore $e^z=o(z^n),\ |z|\to\infty,\ Re(z)<0$

\item[(c)]
$\ds z^n=o(e^{z^2}) \Rightarrow
\lim_{|z|\to\infty}\left|\frac{z^n}{e^{z^2}}\right|=0$ for all
$n>0$

Here $|e^{z^2}|>|z^n|$ for all $z\to\infty$.

Therefore $z^n=o(e^{z^2}),\ z\to\infty$.

\emph{except} on the imaginary axis, when $|e^{z^2}|=1\ \ Re(z)\ne
0$.
\end{description}



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