So I'm trying to show that for $0<a<1$ and for $\epsilon >0$ that $\exp((\log x)^{a})=\mathcal{O}(x^{\epsilon}).$
So this amounts to showing that $\limsup_{x\rightarrow\infty}\dfrac{\exp((\log x)^{a})}{x^{\epsilon}}<\infty$. I tryed to use l'hopital's on $\lim_{x\rightarrow 0}\dfrac{\exp((-\log x)^{a})}{1/x^{\epsilon}}$ but that did not seem to work any ideas on how to do this?