functions that f(n) is not O(g(n)) and g(n) is not O(f(n))

Question

Are there any functions such as f(n) and g(n) that both;

f(n) != O(g(n)) and
g(n) != O(f(n)). 

Are there any functions that fulfills the requir

Answer 1

 f(n)=n and g(n)=n^(1 + sin(x)). 

f(n) is not O(g(n)) and g(n) is not O(f(n)).

Refer http://c2.com/cgi/wiki?BigOh

Answer 2

Consider:

f(n) = 0 if n is odd, else n*n
g(n) = n

Then for odd values g(n) is more than a constant factor bigger than f(n) (and so g(n) is not O(f(n)), while for even values f(n) is more than a constant factor bigger than g(n) (and so f(n) is not O(g(n))).

Observe that f(n) does not have a limit at infinity as n approaches infinity, so in some sense this is a cheap example. But you could fix that by replacing 0, n, n*n with n, n*n, n*n*n.

I think if two non-negative functions have the property that f(n)/g(n) has a (perhaps infinite) limit as n approaches infinity, then it follows that one of them is big-O the other one. If the limit is 0 then f(n) is O(g(n)), if the limit is finite then each is big-O the other, and if the limit is infinite then g(n) is O(f(n)). But I'm too lazy to confirm by writing a proof.