I'm learning about asymptotic analysis and, as a starting point, big and little o definitions.
On the Wikipedia page, http://en.wikipedia.org/wiki/Big_O_notation
further down under the heading for little-$\mathcal{o}$ notation it states
"In this way little-$\mathcal{o}$ notation makes a stronger statement than the corresponding big-$\mathcal{O}$ notation: every function that is little-$\mathcal{o}$ of $g$ is also big-$\mathcal{O}$ of $g$, but not every function that is big-$\mathcal{O}$ g is also little-$\mathcal{o}$ of $g$ (for instance $g$ itself is not, unless it is identically zero near ∞"
This makes sense to me, given the definitions. However, I have an example that seems to be to the contary. In the limit $x\rightarrow0$ the little order of the sin function could be characterized as
$$ \sin(x)=\mathcal{o}(1) $$ i.e., it is always a lot less than 1. According to the arument, this should also be the big-$\mathcal{o}$ of $\sin$ in the same limit. However, in the limit the big $\mathcal{O}$ is
$$ \sin(x)=\mathcal{O}(x) $$
and which as at odds with the relationship between litttle-$\mathcal{o}$ and big-$\mathcal{O}$ in the wiki article.
Where have I went wrong? Is it because saying $\sin(x)=\mathcal{o}(1)$ hasn't characterised the order in terms of a dominant function, but rather picked a constant larger than zero?
Cheers