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Jul 3, 2011 at 12:40 comment added GEdgar $f(n) = o(1)$ as $n \to \infty$ means $\lim_{n\to\infty} f(n) = 0$. No need for $\epsilon$ or $N$. But, of course, the definition of the limit can involve $\epsilon$ and $N$.
Mar 26, 2011 at 6:49 comment added Fabian whether there is $\epsilon$ or $1$ in the denominator does not matter at all...
Mar 26, 2011 at 6:35 comment added NebulousReveal see this: math.osu.edu/~fowler/teaching/handouts/ibl-analysis/…
Mar 26, 2011 at 6:24 comment added sigma.z.1980 doesn't f(n)=o(g(n)) mean $limf(n)=|\epsilon \cdot g(n)|$?
Mar 26, 2011 at 6:12 history answered NebulousReveal CC BY-SA 2.5