limits - Why is $sqrt x in O(log(x))$?

Since $\lim_{x \rightarrow \infty}(\frac {\log(x)}{\sqrt x} ) = 0$, we can conclude that $\log(x) \in o(\sqrt x)$.

This implies that $\sqrt x \in O(\log (x))$. Yet, looking at the graphs, there doesn't seem to be a constant multiple of $\log (x)$ that $\sqrt x$ is always less than, because $\sqrt x$ just keeps getting bigger.

Not sure what I'm missing here, any help is appreciated.

Answer

$f(x)\in o(g(x))$ does not imply that $g(x)\in O(f(x)).$ For example, $x\in o(x^2)$, but surely $x^2 \notin O(x).$ What you might write is that $x^2 \in \omega (x).$ (Some people say that $g(x)\in\omega (f(x))$ if $f(x)/g(x)$ converges to $0$, as $x\to \infty.$)

Loosely speaking, $f\in O(g)$ means that $f/g$ is bounded in the long run, and $f\in o(g)$ means that $f(x)/g(x)$ diminishes to zero as $x\to\infty.$ So you see, $f\in o(g)$ merely implies $f\in O(g)$; $g/f$ can very well be unbounded!