Little 'o' / Big 'O' Definitions

I understand, or at least think I understand, the nature of a function that is "little o": If $f$ is a function between Banach spaces E and F, then it is "little-o" if

$$|x|\rightarrow 0 \implies \frac{|f(x)|}{|x|} \rightarrow 0$$

Thus the evaluation of $f$ at $x$ approaches $0$ faster than $x$ itself. I have read other posts on here, such as this one that give a different definition. Also, textbook authors don't seem to be in agreement either. For instance, in their advanced calculus text, Loomis and Sternberg declare a function to be "little o" if it satisfies essentially the definition I just gave but also add the condition that $f(0) = 0$. On the other hand, Marsden et. al. in "Manifolds, Tensor Analysis and Applications" define a "little o" function as any continuous function $f:E\rightarrow F$ such that $$\lim_{x\rightarrow 0}\frac{f(x^k)}{|x|^k} = 0$$

Is there any hope of reconciling these definitions? They seem to be saying approximately the same thing, but not quite.

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I've never head of a function being simply "little o", period. It is always "little o of something". Namely, $f$ is $o(g(x))$ iff $\frac{|f(x)|}{|g(x)|}\to 0$ for the limit in question. – Henning Makholm Oct 23 '11 at 4:48
@HenningMakholm See, for instance Loomis and Sternberg Advanced Calculus p137; this is how they define "little o" – ItsNotObvious Oct 23 '11 at 5:24
Well, pics or it didn't happen! And if it did happen, I'd be rather wary of trusting that test for anything of importance. – Henning Makholm Oct 23 '11 at 5:33
The book's available at math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf – Gerry Myerson Oct 23 '11 at 6:48
What Loomis and Sternberg actually do is they define a set of functions, which set they denote by a symbol that looks to me like some sort of Gothic little-oh, by saying $f$ is in that family if $f(0)=0$ and $\|f(x)\|/\|x\|\to0$ as $x\to0$. So they never say "$f$ is little-oh;" rather, "$f$ is in little-oh." – Gerry Myerson Oct 23 '11 at 21:53

This is more of a long comment to the comment of Henning Makholm. The objective of the $o$ and $O$ notations is to compare growth (asymptotic behavior) of 2 functions $f$ and $g$. The function $g$ doesn't have to be $x$ or $x^k$. Defining $f(x) = \underset{x\to a}{o}(g(x))$ as saying $|f(x)|/|g(x)| \underset{x\to a}{\to} 0$ is also bad because it leads to writing nonsense when $g$ has zeros.

A good definition would be $f(x) = \underset{x\to a}{o}(g(x))$ if there exists a nonnegative function $\varepsilon$ and a neighborhood $U$ of $a$ such that $\varepsilon(x) \underset{x\to a}{\to} 0$ and for every $x$ in $U\setminus\{a\}$, $|f(x)| = \varepsilon(x)|g(x)|$.

Similarly $f(x) = \underset{x\to a}{O}(g(x))$ if there exists a nonnegative bounded function $\varepsilon$ and a neighborhood $U$ of $a$ such that for every $x$ in $U\setminus\{a\}$, $|f(x)| = \varepsilon(x)|g(x)|$.

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That, and if you want a correct definition, add the condition that $\varepsilon$ is defined on a neighborhood of $a$ minus $\{a\}$ (or that $|f|=\varepsilon\cdot|g|$ on a neighborhood of $a$ minus $\{a\}$). – Did Oct 23 '11 at 17:27
Thanks for pointing that out. – AFK Oct 23 '11 at 17:34
Your revised version was still not quite correct and I took on me to modify it. If you do not like the result, please go back to the previous version. Compare with this. – Did Oct 23 '11 at 17:42
Your edit is fine. I didn't mention explicitly the neighborhood not to be redundant but it is indeed better to make the quantification explicit. – AFK Oct 23 '11 at 20:51