Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that $f(x) = o(g(x))$ for $x \to \infty $ if (and only if) $\lim_{x \to \infty}\frac{f(x)}{g(x)}=0$ Which means than $f(x)$ has a order of growth less than that of $g(x)$.

1) I'm still confused if $x \to 0$. Because in this case $x^5 = o(x^2)$

2) Can someone list me the properties of little-o? For now, I know the following:

$f(x)*o(g(x) = o(f(x)*g(x))$

$o(f(x)) \pm o(f(x)) = o(f(x))$

Thank you!

share|cite|improve this question
Informally, little-o means "much smaller than". So $x^5 = o(x^2)$ as $x \to 0$ is true since $x^5$ is much smaller than $x^2$ when $x$ is close to $0$. – Antonio Vargas Dec 4 '12 at 22:01
thank you Antonio. I also used wolframalpha to plot various functions. Now I have a visual representation of what's happening. – Umar Jamil Dec 4 '12 at 22:08
up vote 4 down vote accepted

Thanks to Wikipedia** here are some properties:

  • $ o(f) + o(f) \subseteq o(f) $
  • $ o(f) o(g)\subseteq o(fg) $
  • $ o(o(f)) \subseteq o(f) $
  • $ o(f) \subseteq O(f) $


Also, the following document may help you:

share|cite|improve this answer
thank you, I already read wikipedia before posting, but I wasn't sure if it listed also the particular cases which are usesful in most cases. My course doesn't include Big-O, so I might get confused reading the definition related to it. Can you please only tell me why the statement I wrote is true? I am unable to imagine a sort of mental representation of what's happening to the function... – Umar Jamil Dec 4 '12 at 21:45
It is your welcome, the document from Caltech is more informative, go ahead with it. – Ragnar Dec 4 '12 at 21:48
Ok, I'll try with it and let you know if I have problems understanding something. – Umar Jamil Dec 4 '12 at 21:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.