Quasi-polynomial time complexity and proof In my algorithm class, we were talking about a function $f(x)$ that grows faster than any polynomial ($x^c$), but is outgrown by any exponential ($c^x$), where $c > 1$. That is, a function $f(x)$, such that both the sequences $P_x = \frac{x^c}{f(x)}$ and $E_x = \frac{f(x)}{c^x}$ converges to $0$, for all $c > 1$.
I'm considering a function $f(x) = x^{log_2x}$, and I've proven the sequence $P_x$ converges to $0$ as $x \to \infty$, but I'm having some trouble proving that $E_x$ also converges to $0$. (I've tried the epislon approach, but it generates really messy algebra that I can't deal with).
In fact, I'm still not a hundred percent sure if $f(x) = n^{log_2n}$ is such a function, I've tried to verify with my calculator that when $c$ is really close to $1$, $\lim_{x \to \infty}\frac{f(x)}{c^x} = 0$. However, as I said, I don't know how to construct a solid proof. Any hint?

Note that $$ n^{\log_c(n)} = \Big( c ^{\log_c (n)} \Big)^{\log_c(n)} =  c^{(\log_c(n))^2}.$$
Since $(\log_c(n))^2 < n$ for large enough $n$, $\lim_{n\to \infty} \frac{n^{\log_c(n)}}{c^n} = 0$ follows.
