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This is the big O-Notation as given today in one of our exercise classes (it is sloppy but just as it was on the blackboard):

$$\begin{align*} &f = O(g):\quad\limsup_{x \rightarrow a}\, \left|\frac{f}{g}\right| < \infty\\\\ &f = o(g):\quad\lim_{x \rightarrow a} \,\left|\frac{f}{g}\right| = 0 \end{align*}$$

I asked the Teaching assistant why in the first case the limit superior is used and in the second only the limit, but he didn't know. A fellow student said that it is maybe because if we have convergence to zero, it follows that lim sup = lim inf. Is this correct? And if so, why?

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Are those $a$s suppose to be $\infty$? – user7530 Feb 19 '13 at 15:51
No, he said not necessarily. He said "usually "a" is zero or +/- infinity" – user62487 Feb 19 '13 at 15:52
up vote 2 down vote accepted

To say that $f$ is $O(g)$ is to say that there is a positive constant $c$ such that $|f(x)|\le c|g(x)|$ for all $x$ sufficiently close to $a$. This just imposes a kind of upper bound on $|f(x)|$ near $a$; it doesn’t say anything about how small $|f(x)|$ can be. Thus, it only limits $\sup|f(x)|$ as $x\to a$; $|f(x)|$ can oscillate wildly as long as it never gets bigger than $c|g(x)|$ in some neighborhood of $a$.

When you say that $f$ is $o(g)$, on the other hand, you’re saying that for every $c>0$ there is a neighborhood of $a$ in which $|f(x)|\le c|g(x)|$, which is precisely equivalent to saying that you can make $\left|\frac{f(x)}{g(x)}\right|$ as small as you like by taking $x$ close enough to $a$. But this is exactly what it means to say that $\lim_{x\to a}\left|\frac{f(x)}{g(x)}\right|=0$.

If this is your first exposure to big-O and little-o notation, you may find it useful to see the more common definitions here.

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Matches my intuition, thanks a lot, also for the link. – user62487 Feb 19 '13 at 16:08
@user62487: You’re welcome. – Brian M. Scott Feb 19 '13 at 16:13

Well, just because $f = o(g)$ means that $f$ is so small compared to $g$, $limsup$ and $liminf$ would have the same value. Thus, $f = o(g)$ means all limits ($lim$, $liminf$ and $limsup$) will converge to same value.

When you are talking about big-O, all we can say is that $\left|\frac{f}{g}\right|$ will not exceed some number $M$. But hey, this ratio can have different $sup$ and $inf$ values, so we are talking only about biggest value of them.

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