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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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    Isn't the second just spelling out what the limit means?2011-03-06
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    @Fabian: I am not sure if the limit of the ratio always exist. For example, if no matter how close you approach $\infty$, can there always exist $x$ closer to infty so that $g(x)=0$? Similar question for f∈ω(g) and for f∼g?2011-03-06
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    Now I see what you mean. But what if you replace lim with limsup (wikipedia is not always the best place to get the definitions).2011-03-06
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    In that spirit I would also say that there is an absolute value missing. What if we would rewrite the definition in Wikipedia (we have this power ;-)) and define little-o as $\limsup_{x\to\infty} \left| \frac{f(x)}{g(x)} \right| = 0$?2011-03-06
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    @Fabian: (1) If replace lim with limsup, is taking absolute value of the ratio needed for the two definitions to be equivalent? (2) for f∈ω(g), do you also suggest to replace lim with liminf? how about for f∼g (3) Do you mind post the definition and its source that are the best you can find?2011-03-06
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    I don't have really a reference at hand. I think I got confused a bit. You two definitions are in fact equivalent. The second is just writing out what limit means. If $g(x)$ is zero infinitely often, then neither the limit is zero nor can you find an $x_0$ for every $M$.2011-03-06
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    @Fabian: Thanks! But I still don't understand if the two are equivalent. If g is zero infinitely often: then o(g) in the first definition will be empty; and o(g) in the second will be nonempty, where some f's are 0 whenever g is 0, although I am not sure if f being constantly 0 is the only element in o(g)?2011-03-06
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    Take $f=g=\sin(x)$. The $g$ and $f$ are zero infinitely often. And $f= o(g)$ in both definitions. So $o(g)$ is not empty in the first definition...2011-03-06
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    @Fabian: Thanks! Is f(x)/g(x) needed to be well-defined when x is sufficiently large, for its limit as x goes to infinity to be meaningful? If yes, g cannot be in the denominator.2011-03-06
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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 $\lim_{x\to\infty}f(x)/g(x) = 0$.