I know it evaluates to $\frac{rt-\frac{1}{2}\sigma^2t}{2\sigma\sqrt{t}}$ but how to get there is the problem.
Using L'Hôpital you find $$\lim_{n\rightarrow\infty}\frac{e^{rt/n+\sigma\sqrt{t/n}}-1}{e^{2\sigma\sqrt{t/n}}-1} = \frac{1}{2}$$But L'Hôpital doesn't work on the whole thing. The only limit calculator that could figure it out was wolfram but that couldn't give me the steps to get there. using Taylor series expansion gives me:$$\lim_{n\to\infty}\sqrt{n}(\frac{\sum_{k=0}^{\infty}(rt/n+\sigma\sqrt{t/n})-1}{\sum_{k=0}^{\infty}(2\sigma\sqrt{t/n})^k-1}-\frac{1}{2})$$But I fail to see how to go further from here, any help would be appreciated
$\displaystyle n:=\frac{t}{x^2}>0$
$$\lim\limits_{n\rightarrow \infty} \sqrt{n}\left(\frac{e^{rt/n+\sigma\sqrt{t/n}}-1}{e^{2\sigma\sqrt{t/n}}-1}-\frac{1}{2}\right)= \lim\limits_{x\rightarrow 0} \frac{\sqrt{t}}{2\sigma x}\left((rx+\sigma)\frac{e^{rx^2+\sigma x}-1}{rx^2+\sigma x}\frac{2\sigma x}{e^{2\sigma x}-1}-\sigma\right)$$
$$= \lim\limits_{x\rightarrow 0} \frac{\sqrt{t}}{2\sigma} (r+(rx+\sigma)\frac{rx+\sigma}{2!}\frac{(2\sigma x)^0}{0!}B_0+(rx+\sigma)\frac{(rx^2+\sigma x)^0}{1!}\frac{2\sigma }{1!}B_1+x\cdot …)$$
$$=\frac{\sqrt{t}}{2\sigma} (r-\frac{\sigma^2}{2})$$
$B_n$ are the Bernoulli numbers and we need here $B_0=1$ and $\displaystyle B_1=-\frac{1}{2}$ .