<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
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\begin{document}


{\bf Question}

Find the orders of the following expressions as $\varepsilon \to
0^+$

\begin{description}
\item[(a)]
$\ds \frac{1-\cos\varepsilon}{1+\cos\varepsilon}$

\item[(b)]
$\ds \frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}$

\item[(c)]
$\rm{sech}^{-1}\varepsilon$

\item[(d)]
$\ds e^{\tan\varepsilon}$

\end{description}

\vspace{.5in}

{\bf Answer}

The trick is to \lq\lq\ guess" (by expanding as $\varepsilon\to
0^+$) the answer and substitute back in to prove it.

\begin{description}

\item[(a)]
$\ds \frac{1-\cos\varepsilon}{1+\cos\varepsilon} \rightarrow
\frac{1-\left(1-\frac{\varepsilon^2}{2}+\cdots\right)}{1+\left(1+\frac{\varepsilon^2}{2}+\cdots\right)}
\approx \frac{\varepsilon^2}{4}+\cdots$

Also $\ds \left|\frac{1-\cos\varepsilon}{1+\cos\varepsilon}\right|
=\frac{1-\cos\varepsilon}{1+\cos\varepsilon}\ \ \varepsilon\to
0^+$

Therefore
\begin{eqnarray*} \ds \lim_{\varepsilon\to 0^+}
\frac{\left|\frac{1-\cos\varepsilon}{1+\cos\varepsilon}\right|}{|\varepsilon^2|}
&amp; = &amp; \lim_{\varepsilon\to 0^+}
\frac{1-\cos\varepsilon}{\varepsilon^2(1+\cos\varepsilon)}\\ &amp; &amp;
\textrm{by\ l'H\^opital}\\ &amp; = &amp; \lim_{\varepsilon\to
0^+}\frac{\sin\varepsilon}{2\varepsilon(1+\cos\varepsilon)-\varepsilon^2\sin\varepsilon}\\
&amp; = &amp; \lim_{\varepsilon\to
0^+}\frac{\cos\varepsilon}{2(1+\cos\varepsilon)-4\varepsilon\sin\varepsilon
-\varepsilon^2\cos\varepsilon}\\ &amp; = &amp; \frac{1}{4}&gt;0
\end{eqnarray*}

$\ds \Rightarrow
\frac{1-\cos\varepsilon}{1+\cos\varepsilon}=O(\varepsilon^2)\ \
\varepsilon\to 0^+\ \ \ \left(k \geq \frac{1}{4}\right)$

\item[(b)]
$\ds \frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}
\rightarrow
\frac{\varepsilon^{\frac{3}{2}}}{1-1+\frac{\varepsilon^2}{2}+\cdots}
\approx
2\frac{\varepsilon^{\frac{3}{2}}}{\varepsilon^2}=\frac{2}{\varepsilon^{\frac{1}{2}}}\
\ \ \varepsilon \to 0^+$

Also $\ds
\left|\frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}\right|
=\frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}\ \ \
\varepsilon\to 0^+$

Therefore
\begin{eqnarray*}
\lim_{\varepsilon\to 0^+}
\frac{\left|\frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}\right|}
{|\frac{1}{\varepsilon^{\frac{1}{2}}}|} &amp; = &amp; \lim_{\varepsilon\to
0^+}\frac{\varepsilon^2}{1-\cos\varepsilon}\\ &amp; = &amp;
\lim_{\varepsilon\to 0^+} \frac{2\varepsilon}{\sin\varepsilon}\\ &amp;
= &amp; \lim_{\varepsilon\to 0^+} \frac{2}{\cos\varepsilon}\\ &amp; = &amp;
2&gt;0
\end{eqnarray*}

$\ds \Rightarrow
\frac{\varepsilon^{\frac{3}{2}}}{1-\cos\varepsilon}=O(\varepsilon^{-\frac{1}{2}})\
\ \varepsilon\to 0+\ \ \ \left(k \geq 2\right)$

\item[(c)]
Let ${\rm sech}^{-1}\varepsilon=u$ so that ${\rm
sech}u=\varepsilon\ \Rightarrow \cosh u=\frac{1}{\varepsilon}$

PICTURE (both graphs) \vspace{1in}

$\ds \Rightarrow e^u+e^{-u}=\frac{2}{\varepsilon}$. Now as
$\varepsilon \to 0,\ u \to \infty$ so we have

$\ds e^u \approx \frac{2}{\varepsilon}$ or $u \approx \ln 2+\ln
\left(\frac{1}{\varepsilon}\right) \Rightarrow
\rm{sech}^{-1}\varepsilon \approx \ln
2+\ln\left(\frac{1}{\varepsilon}\right)\ \ \varepsilon\to 0$

So try $\ds {\rm
sech}^{-1}\varepsilon=O\left(\ln\left(\frac{1}{\varepsilon}\right)\right)$:

\begin{eqnarray*} \lim_{\varepsilon\to 0^+}
\left|\frac{\rm{sech}^{-1}\varepsilon}{\ln(\frac{1}{\varepsilon})}\right|
&amp; = &amp; \lim_{\varepsilon\to
0^+}\frac{\rm{sech}^{-1}\varepsilon}{\ln(\frac{1}{\varepsilon})}\\
&amp; = &amp; \lim_{\varepsilon\to 0^+} \frac{1}{-\frac{1}{\varepsilon}}
\times -\frac{1}{\varepsilon\sqrt{1-\varepsilon^2}}\\ &amp; = &amp;
\lim_{\varepsilon\to 0^+} \frac{1}{\sqrt{1-\varepsilon^2}}\\ &amp; = &amp;
1&gt;0 \end{eqnarray*}

$\ds \Rightarrow {\rm
sech}^{-1}\varepsilon=O\left(\ln\left(\frac{1}{\varepsilon}\right)\right)\
\ {\rm as}\ \varepsilon\to 0^+$

\item[(d)]
$e^{\tan\varepsilon} \approx e^{\varepsilon+\cdots}
\approx1+\varepsilon+\cdots\ {\rm as} \ \varepsilon\to 0^+$

Therefore try $O(1)$.

$|e^{\tan\varepsilon}|=e^{\tan\varepsilon}\ {\rm as} \
\varepsilon\to 0^+$

Therefore $\ds \lim_{\varepsilon\to
0^+}\left|\frac{e^{\tan\varepsilon}}{1}\right|=\lim_{\varepsilon\to
0^+} e^{\tan\varepsilon}=1$ (regular limit)

$\Rightarrow e^{\tan\varepsilon}=O(1),\ \varepsilon\to 0^+\ \ \ (k
\geq 1)$

\end{description}



\end{document}
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