How is this term derived?

Question

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?

Answer 1

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 $o(\delta)$.