<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}
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\newcommand{\undb}{\underbrace}
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\begin{document}


{\bf Question}

For small real positive $\varepsilon$ show that

\begin{description}
\item[(a)]
$\sinh\left(\frac{1}{\varepsilon}\right)=O\left(e^{\frac{1}{\varepsilon}}\right)$

\item[(b)]
$\log(1+\sin\varepsilon)=O(\varepsilon)$

\item[(c)]
$\log(2+\sin\varepsilon)=O(1)$

\end{description}

\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\rm{sinh}\left(\frac{1}{\varepsilon}\right) =
\frac{e^{\frac{1}{\varepsilon}}-e^{-\frac{1}{\varepsilon}}}{2}$.

Therefore

$\ds \lim_{\varepsilon\to 0^+}
\left|\frac{\rm{sinh}\left(\frac{1}{\varepsilon}\right)}{e^{\frac{1}{\varepsilon}}}\right|
 =  \lim_{\varepsilon\to 0}
\frac{e^{\frac{1}{\varepsilon}}-e^{-\frac{1}{\varepsilon}}}{2e^{\frac{1}{\varepsilon}}}
 =  \lim_{\varepsilon\to
0}\frac{1}{2}-\frac{1}{2}e^{-\frac{2}{\varepsilon}} \frac{1}{2}&gt;0$

$\ds \Rightarrow \sinh
\left(\frac{1}{\varepsilon}\right)=O\left(\frac{1}{\varepsilon}\right),\
\varepsilon\to 0^+$


\item[(b)]
$\log(1+\sin\varepsilon)&gt;0$ as $\varepsilon\to 0^+$

Therefore

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

$\Rightarrow \log(1+\sin\varepsilon)=O(\varepsilon)\ \
\varepsilon\to 0^+$

\item[(c)]
$\log(2+\sin\varepsilon)&gt;0$ as $\varepsilon\to 0^+$

Therefore

\begin{eqnarray*} \lim_{\varepsilon\to 0^+}
\left|\frac{\log(2+\sin\varepsilon)}{1}\right| &amp; = &amp;
\lim_{\varepsilon\to
0^+}\left[\frac{\log(2+\sin\varepsilon)}{1}\right]\\ &amp; = &amp; \log
2&gt;0 \end{eqnarray*}

$\Rightarrow \log(2+\sin\varepsilon)=O(1)$ as $\varepsilon\to 0^+$


\end{description}



\end{document}
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