What can we conclude from "f is not little-o of g"? Given two functions $f$ and $g$, what does "$f$ is not $o(g)$" mean ? What can we conclude from this statement ?
I know "$f$ is $o(g)$" means the limit at infinity of $\frac fg$ is zero.
So does "$f$ is not $o(g)$" mean the limit at infinity of $\frac fg$ is different from zero or does it mean that this limit doesn't exist ?
 A: $f = o(g)$ does mean that the $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$. Notice that saying $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$ is actually the statement "the limit of $\frac{f(x)}{g(x)}$ exists as $x \to \infty$ AND the limit is $0$". The negation of this, "$f$ is not $o(g)$", means that we negate the previous statement resulting in "the limit does not exist OR if it does exist, it is not $0$."
A: That $f$ is not $o(g)$ simply means that there is some $\epsilon>0$ such that for any $N$ there is an $x>N$ for which $|f(x)|>\epsilon|g(x)|$. 
Note that this formulation does not require $g$ to be nonzero for $x$ large enough. If the set of $x$ where $g(x)$ is zero is unbounded, clearly $f(x)/g(x)$ does not converge as $x\to  \infty$, since there is an unbuonded set of points where the fraction is undefined. 
But even if $g(x)\ne 0$ for $x$ large, the formulation in the first paragraph is not strong enough to allow us to conclude anything about whether the limit exists. For a silly example, if $g(x)=2$ for all $x$, and $f(x)=2$ for $x$ an integer, but $f(x)=0$ otherwise, then $f$ is not little-oh of $g$, but the limit does not exist (in this case, we can take $\epsilon=1$). On the other hand, if $g(x)=2$ for all $x$, and $f(x)=4$, then the limit exists.
By the way, the big-oh notation is due to Bachmann (around 1894), and later adopted by Landau (to whom many attribute it), and it was already common in the early twentieth century. On the other hand, the little-oh notation, due to Landau (around 1909) took a bit longer to be adopted. For example, Hardy's book Orders of infinity uses the $O$ notation, but does not use $o$, and uses $\epsilon$ instead of $o(1)$, as in 
 $$ n!(e/n)^n=(1+\epsilon)\sqrt{2\pi n}. $$ 
