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  1. Let f(x) and g(x) be two functions defined on some subset of the real numbers. There are two definitions for $f \in o(g)\mbox{ as }x\to\infty\,$, according to Wikipedia:

    • $\lim_{x \to \infty}\frac{f(x)}{g(x)}=0.$
    • for every $M > 0$, there exists a constant $x_0$, such that $|f(x)| \le \; M |g(x)|\mbox{ for all }x>x_0. $

    I was wondering if the two definitions are equivalent? Can it be possible that the second one is more general than the first one in that the limit of the ratio may not exist?

  2. Similar questions for $f \in \omega(g)$ and for $f \sim g$?

Thanks and regards!

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    I guess it is a bit a matter of whom you ask (the next sentences are quite vague). In order that the second definition hold $g$ and $f$ need to have the zeros at the same places. These are then places where $f/g$ is undefined but can be assigned an unique value such that $f/g$ becomes continuous. Essentially, you can remove the poles from $f/g$. All this thing whoever in practice play no role as $g$ typically is $x^\alpha$ or $\log(x)$ or something like that...2011-03-06

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The two definitions of $“f(x)\in o(g(x)) \mbox{ as } x\to\infty”$ you stated are equivalent as long as $g(x)\neq0$ for every sufficiently large $x$. Usually we use the $O$-notation and its relatives only when this condition holds. Under this condition, the second definition is just what we mean by the first definition.

If this condition is not satisfied, then there are arbitrarily large points $x$ at which the function $f(x)/g(x)$ is not defined. In this case, it is not clear what we mean by $\lim_{x\to\infty}f(x)/g(x) = 0$.