Osculating plane<=>distance limit Let $r$ be a unit-speed bi-regular curve. (It passes the point $s_0$)
Let $distP(q)$ be the distance between the plane $P$ and the point $q$.
Question.
The plane is equal to the osculating plane of $r$ at $s_0$ if and only if
$P$ contains 
$r(s_0)$,$$ \lim_{s \to s_0} \frac{distP(r(s))}{(s-s_0)} = 0$$
$$\lim_{s \to s_0} \frac{distP(r(s))}{(s-s_0)^2} =0$$

I found the answer. Using the L'Hospital's law, the equation means that the binormal vector of r(s) perpendicular to the normal vector. And it is true because it is bi-regular curve.
