How is this term derived? The following screenshot is taken from some of my lecture notes, and attempts to prove that
$$
\frac{\exp(r \delta) - \exp(\mu \delta - \sigma \sqrt{\delta})}{\exp(\mu \delta + \sigma \sqrt{\delta}) - \exp(\mu \delta - \sigma \sqrt{\delta})}
$$

I am struggling to see where the circled term came from here. I understand that the first bracket came by using the (second) limit rule provided, with $\delta \mapsto 2 \sigma \sqrt{\delta}$, however, I can't see how the second bracket of this term came about.
Can anyone help me to understand this?
 A: We recall that per definition
\begin{align*}
f(\delta)=o(g(\delta))\qquad \delta\to 0
\end{align*}
means
\begin{align*}
\lim_{\delta\to 0}\frac{f(\delta)}{g(\delta)}=0
\end{align*}
It follows for $\delta\to 0$
\begin{align*}
\delta^{\alpha}=o(\delta)\qquad \alpha>1\tag{1}
\end{align*}
which means that $o(\delta)$ swallows all powers of $\delta$ with exponent greater one.

Expanding the exponential series gives according to (1)
  \begin{align*}
\exp(\delta)-1&=\delta+\frac{1}{2}\delta^2+o(\delta)\tag{2}\\
&=\delta+o(\delta)
\end{align*}

It is sufficient to consider the numerator inside the second bracket.

We obtain according to (2)
  \begin{align*}
\color{blue}{\exp}&\color{blue}{(r\delta-\mu\delta+\sigma\sqrt{\delta})-1}\\
&=r\delta-\mu\delta+\sigma\sqrt{\delta}+\frac{1}{2}\left(r\delta-\mu\delta+\sigma\sqrt{\delta}\right)^2+o(\delta)\\
&=r\delta-\mu\delta+\sigma\sqrt{\delta}\\
&\qquad+\frac{1}{2}\left(r^2\delta^2+\mu^2\delta^2+\sigma^2\delta-2r\mu\delta^2+2r\sigma\delta\sqrt{\delta}-2\mu\sigma\delta\sqrt{\delta}\right)+o(\delta)\tag{3}\\
&\,\,\color{blue}{=r\delta-\mu\delta+\sigma\sqrt{\delta}+\frac{1}{2}\sigma^2\delta+o(\delta)}
\end{align*}
  whereby terms $\delta^{\alpha}$ with $\alpha>1$ from (3) are swallowed by $o(\delta)$.

