# Big-O notation question

We say that $f(x) = O(g(x))$ if there exists a constant $C$ such that $|f(x)| \leq C|g(x)|$ for all $x \in \mathbb{R}$ (or whatever we take our domain to be). What is meant explicitly by the statement that $f(x)$ is not $O(g(x))$?

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It is more usual to require $|f(x)|\le C|g(x)|$ only for $x$ in some (punctured) neighborhood of the limit we're considering. For example if the limit in $x\to+\infty$ then there must be $N$ and $C$ such that $|f(x)|\le C|g(x)|$ for $x>N$, and if it is $x\to 0$, then there must be $\varepsilon$ such that it holds for $|x|<\varepsilon$, $x\ne 0$. Otherwise we couldn't say that $\log x = O(x)$ for $x\to\infty$, for example. –  Henning Makholm Nov 13 '11 at 0:25

For any $C$, there exists a value of $x$ such that $|f(x)| \gt C|g(x)|$.