Let us think ${s}$ is complex values as ${s=a+ib}$
And see that with $x_0=\pi$ with $u>1$ as $u\in\mathbb{R}$
$$\sin(x_0+\frac{s}{u})\neq0$$
Now, could we apply L'hospital rule to the following function? Why ?
$$\lim_{u \to \infty}\frac{\sin 2(x_0+\frac{s}{u})}{\sin(x_0+\frac{s}{u})} $$
Because we cannot apply L'hospital rule to the following one:
$$\lim_{x \to 0}\frac{x}{(x+x^2e^{i/x^2})}=1 $$
L'Hopital requires the mean value theorem which holds in general only for real valued functions defined on real intervals; now for complex valued functions one can sometimes come up with something that is working using Taylor series.
However the second example is oscillating badly which is reflected in the derivative so it's easy to see that any useful analogue of MVT doesn't hold; in other words if $f(x)=x+x^2e^{i/x^2}$ then $f(x)-f(0)=f(x)$ is not in general $xf'(t_x)$ for any $|t_x| \to 0$
For the first example you can argue as follows: $$\lim_{u \to \infty}\frac{\sin 2(x_0+\frac{s}{u})}{\sin(x_0+\frac{s}{u})}=\lim_{v \to 0}\frac{\sin 2(x_0+sv)}{\sin(x_0+sv)}$$ and using the Taylor series we have $\sin 2(x_0+sv)=2sv \cos 2(x_0+c(v)), c(v) \to 0, v \to 0$ and same for denominator.
So in general applying L'Hopital outside its original formulation (real functions on real intervals) requires looking at the specific case and arguing by its properties.