Domain of attraction $F(x)=\exp(-x-\sin(x))$

I need to show that $$F(x)=\exp(-x-\sin(x)),x >0$$ is in no domain of attraction. That means that there exist no $$a_{k} >0, b_{k} \in \mathbb{R}, k \in \mathbb{N}$$ with $$\lim\limits_{k \to \infty} \frac{U(kx)-b(k)}{a(k)}=D(x)$$ where $$D(x)=G^{\leftarrow}(e^{\frac{1}{x}}),$$ $$G$$ is a nondegenerate distribution function and $$U=(\frac{1}{1-F})^{\leftarrow}$$. As a hint i know that the following holds $$\lim\limits_{k \to \infty} U(n_{k}x)-\log(n_{k})= U_{1}(x)$$ where $$U_{1}$$ is the inverse of $$\exp(x+\sin(x))$$ and $$n_{k}=[\exp(2\pi k)].$$ Any approach?