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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
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\begin{document}


{\bf Question}

Is it true that $\{l+o(1)\}\cosh x-\{1+o(1)\}\sinh
x=\{1+o(1)\}e^{-x}$, as $x \to \infty$?

\vspace{.5in}

{\bf Answer}

No!!

\begin{eqnarray*} &amp; &amp; [1+o(1)]_{(1)}\cosh x-[1+o(1)]_{(2)}\sinh x\\
 &amp; = &amp; \frac{1}{2}\left\{[1+o(1)]_{(1)}e^x+[1+o(1)]_{(1)}e^{-x}\right\}
 \\ &amp; &amp; -\frac{1}{2}\left\{[1+o(1)]_{(2)}e^x-[1+o(1)]_{(2)}e^{-x}\right\}\\
 &amp; = &amp;
 \frac{1}{2}[o(1)_{(1)}-o(1)_{(2)}]e^x\\ &amp; &amp; +\frac{1}{2}
 \left\{[1+o(1)]_{(1)}+[1+o(1)]_{(2)}\right\}e^{-x}\\
 &amp; = &amp; o(e^x), \end{eqnarray*}

since $o(1)_{(1)} \neq o(1)_{(2)}$ necessarily,

e.g., $\ds \frac{1}{x} \neq \frac{1}{x^2}$ as $x \to +\infty$ so
it \emph{doesn't} vanish.

\bigskip

$(\star)$ To show $o(1)-o(1)=o(1)$, let $f(x)=o(1),\ g(x)=o(1)$.

Then $f(x) \leq K_f,\ g(x) \leq K_g\ \ \ K_f,\ K_g&gt;0,\ x \to
+\infty$

Then \begin{eqnarray*} |f(x)-g(x)| &amp; \leq &amp; |f(x)|+|g(x)|\ {\rm
by\ triangle\ inequality}\\ &amp; \leq &amp; K_f+K_g=K\ {\rm say\ as\ }\ x
\to +\infty \end{eqnarray*}

Therefore $|f(x)-g(x)| \leq K \Rightarrow f-g=o(1) \Rightarrow
o(1)-o(1)-o(1)$

Similarly $o(1)+o(1)=o(1)$.

\bigskip

$(\star\star)$ To show $o(1)e^x+[1+o(1)]e^{-x}=o(e^x)$

Clearly $e^{-x}=o(e^x)\ {\rm as} \ x \to +\infty$

Therefore we must show that $o(1)e^x =o(e^x)\ \ \ x \to +\infty$

Let $\ds h(x)=o(1),\ \ h&gt;0,\ x \to \infty \Rightarrow \lim_{x \to
\infty} h(x)=0$

Therefore $\ds \lim_{x \to \infty} \frac{h(x)e^x}{e^x}=\lim_{x \to
\infty}h(x)=0\Rightarrow o(1)e^x=o(e^x)$




\end{document}
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