I don't immediately see how to apply L'hopitals rule directly to the given limit, but I have an approach where it shows up in the end if you wish.
By conditioning on $\mathcal{F}_{n-1}$, we have
$$
\mathbb{E}\left(e^{iuS_n/\sqrt n}\right) = \mathbb{E}\left(\mathbb{E}\left(e^{iuS_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right)\right) = \mathbb{E}\left(e^{iuS_{n-1}/\sqrt n}\mathbb{E}\left(e^{iuX_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right)\right).
$$
To use the assumptions on $X_n$ which are given, we use a Taylor expansion of the exponential. This gives us that there is some constant $C > 0$ such that
$$
\left|\mathbb{E}\left(e^{iuX_n/\sqrt n}\middle|\mathcal{F}_{n-1}\right) - \mathbb{E}\left(1 + \frac{iu}{\sqrt n}X_n - \frac{u^2}{2n}X_n^2\middle|\mathcal{F}_{n-1}\right)\right| \leq \frac{C}{n^{3/2}}\mathbb{E}(|X_n^3||\mathcal{F}_{n-1}).
$$
If we now use the assumptions on $X_n$ and put everything together, we find that
$$
\left|\mathbb{E}\left(e^{iuS_n/\sqrt n}\right) - \left(1 - \frac{u^2}{2n}\right)\mathbb{E}\left(e^{iuS_{n-1}/\sqrt n}\right)\right| \leq \frac{C}{n^{3/2}}\mathbb{E}(|X_n^3||\mathcal{F}_{n-1}) \leq \frac{CK}{n^{3/2}}.
$$
Repeating this and using the triangle inequality, we obtain that
$$
\left|\mathbb{E}\left(e^{iuS_n/\sqrt n}\right) - \left(1 - \frac{u^2}{2n}\right)^n\right| \leq \frac{CK}{n^{1/2}}.
$$
Since the upper bound goes to 0, we find that
$$
\lim_{n\to\infty} \mathbb{E}\left(e^{iuS_n/\sqrt n}\right) = \lim_{n\to\infty} \left(1 - \frac{u^2}{2n}\right)^n = e^{-u^2/2}.
$$
Here, the latter is a standard limit, but you can compute it using L'hopitals rule by first taking the logarithm.
EDIT: To see how we can use L'hopital's rule to compute the last limit, consider the function $f(x) = x\log\left(1 - \frac{u^2}{2x}\right)$. The limit we need to compute is
$$
\lim_{n\to\infty} e^{f(n)} = e^{\lim_{n\to\infty} f(n)},
$$
where we used the continuity of the exponential. Now
$$
\lim_{n\to\infty} f(n) = \lim_{x\to\infty} x\log\left(1 - \frac{u^2}{2x}\right) = \lim_{x\to\infty} \frac{\log\left(1 - \frac{u^2}{2x}\right)}{1/x}
$$
By L'hopital's rule, this is equal to
$$
\lim_{x\to\infty} \frac{u^2/(2x^2)}{1 - u^2/(2x)}\cdot\frac{1}{-1/x^2} = \lim_{x\to\infty} \frac{-u^2}{2(1 - u^2/(2x))} = -\frac{u^2}{2}.
$$