# Little o(h) limit about h=0

I understand that generally if a function $$f(h)$$ is described as $$o(h)$$ that $$f(h)$$ has a smaller rate of growth than $$h$$ (like it would have to be $$\sqrt{h}$$). i.e. $$\sqrt{h} = o(h)$$, just like (for example) $$4h=o(h^2)$$. The notes I'm reading (CT3), however, states that:

A function $$f(h)$$ is described as $$o(h)$$ if:

$$\lim_{h \to 0} \frac{f(h)}{h} = 0$$

but if I use, for example, $$f(h)= \sqrt{h}$$ which does have a slower growth rate then $$h$$ then the limit doesn't go to $$0$$. Is there a different meaning to little $$0$$ when its approaching $$0$$ compared to when it goes to infinity cause the only way that limit holds is if $$f(h)$$ goes to $$0$$ faster then $$h$$ does which I guess means $$f(h)$$ decreases faster then $$h$$ as $$h \to 0$$.

Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.

• It is a definition - I don't understand what you mean by explanation? You could interpret it as saying $f(0) = 0$ and $f'(0) = 0$. Jan 31, 2015 at 6:51
• Im talking about little o notation? Jan 31, 2015 at 7:06
• I understand. I don't understand what you are asking. Jan 31, 2015 at 7:09
• I just need an explanation of the grey bit. Like how does o(h) work when h->0 instead of going to infinity Jan 31, 2015 at 7:24
• I don't know what you mean, the definition only involves $h \to 0$? It is exactly equivalent to $f(0) =0$ and $f'(0) = 0$. Jan 31, 2015 at 7:28

Yes, the little-o notation (and Landau symbols in general) behaves differently for $x\to 0$ and for $x\to\infty$.

When we're considering $x\to\infty$ (as you may be familiar with from analysis of algorithms), $\sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $\sqrt x = o(x)$.

On the other hand, if we're considering $x\to 0$ (which is more common in analysis), then when $x$ is close to zero, $\sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(\sqrt x)$.

If would be less confusing to make it explicit which limit we're working with, and write something like $$\sqrt x = \mathop o_{x\to\infty}(x) \qquad\qquad\qquad x = \mathop o_{x\to 0}(\sqrt x)$$ However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.

• This is confusing to me. Basically we are using equal notation for two almost opposite behaviour. Sep 27, 2022 at 21:28
• I agree to @Curiousstudent. How can this be true, and never mentioned anywhere? I checked Wikipedia, MathWorld, a number of online lecture notes ... Is this supposed to be obvious? Very strange ... :) May 11 at 14:58

Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(\sqrt{h})$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)\Leftrightarrow\lim_{x\to x_0}\frac{f(x)}{g(x)}=0$$

• "$\sqrt{h}$ has slower growth than $h$" is for $h \to \infty$, and not for $h \to 0$ as we have here. Oct 4, 2017 at 23:46