3 ms·
Yes. (log n)/(n^epsilon) tends to zero for any positive epsilon, and so log n = n^(o(1)). The same holds for (log n)^k.
by mauricioc 4y ago
Yes. (log n)/(n^epsilon) tends to zero for any positive epsilon, and so log n = n^(o(1)). The same holds for (log n)^k.