My textbook (Introduction to Algorithms) states the following:
When polynomially comparing $n^\epsilon$ and $lgn$, it states that $n^\epsilon$ is polynomially greater for any positive $\epsilon$. Does it mean for positive integers? Because in the case of $\epsilon = 0.1$, we have $n^{0.1}$ compared against $lgn$. Is $lgn$ not polynomially larger in that case?
Also, nowhere that I see in the book elsewhere does it state to assume $\epsilon$ is an integer. In fact, above that there's an example where $\epsilon$ is 0.2, so I don't think that assumption is being made.
Should it read that $\epsilon$ is an integer? Or am I not understanding polynomial comparisons?
Expanding on my comment: take the limit $$\lim_{x\to\infty} \frac{x^\epsilon}{\log x} = \lim_{x\to\infty} \epsilon x^\epsilon = \infty.$$
Therefore for some $n$, $\frac{x^\epsilon}{\log x} > 1$ for $x>n$, so for $x > \max(1,n)$, $$x^\epsilon > \log x.$$