The following is an excerpt from Hull's book "Options, Futures and Other Derivatives (9th edition)", page 299:
$$p^{*}=\left(\frac{e^{r T / n}-e^{-\sigma \sqrt{T / n}}}{e^{\sigma \sqrt{T / n}}-e^{-\sigma \sqrt{T / n}}}\right)\left(\frac{e^{\sigma \sqrt{T / n}}}{e^{r T / n}}\right)\tag{13a.6} $$
By expanding the exponential functions in a series we see that, as $n$ tends to infinity, $p^{*}\left(1-p^{*}\right)$ tends to $1 / 4$ and $\sqrt{n}\left(p^{*}-1 / 2\right)$ tends to $$ \frac{\left(r+\sigma^{2} / 2\right) \sqrt{T}}{2 \sigma}\tag{13a.7} $$
Could someone explain why $p^*$ tends to $\frac{1}{2}$ (or rather, why $p^*\times(1-p^*)$ tends to $\frac14$), and why $\sqrt{n}\left(p^{*}-1 / 2\right)$ tends to the equation noted as $n$ approaches infinity?
$e^x = 1 + x + x^2/2 + x^3/3!...$
This is from the most excellent blog post here - Taylor series
(Admittedly I do not understand the difference between Taylor and maclauren from it, but that seems unimportant).
Now in our problem, we have:
$p^{*}=\left(\frac{e^{r T / n}-e^{-\sigma \sqrt{T / n}}}{e^{\sigma \sqrt{T / n}}-e^{-\sigma \sqrt{T / n}}}\right)\left(\frac{e^{\sigma \sqrt{T / n}}}{e^{r T / n}}\right)\tag{13a.6}$
$\frac{e^{r T / n}-e^{-\sigma \sqrt{T / n}}}{e^{\sigma \sqrt{T / n}}-e^{-\sigma \sqrt{T / n}}} $
Which can boil down to
So we have: