"Let $\alpha(s)$ be a regular curve. Verify that the the binormal vector $b(s_0)$ is the limiting position of the perpendicular to the tangent lines to $\alpha$ at $s_0$ and $s_1$ as $s_1$ tends to $s_0$."
I'm not really getting what the author means here. Am I supposed to prove that as $s_0$ approaches $s_1$, the plane determined by $\alpha(s_0)$ and the perpendicular line at $\alpha(s_1)$ approaches the osculating plane?
EDIT: Still no closer to figuring out precisely what "limiting position of the perpendicular the tangent lines" really means. Would be grateful for any help.
Thanks to @TedShifrin (whose patience was crucial for this), the exercise is solved and all my initial confusions are cleared.
Let $s_1=s_0+h$. The intersection line of the normal planes at $\alpha(s_0)$ and $\alpha(s_1)$ has direction vector $T(s_0)\times T(s_0+h)$. As $h\to 0$, it's clear that $\alpha(s_0)$ will be a point in the limiting intersection. The \limiting direction vector (defined up to scalar multiples) will be \begin{align*} \lim_{h\to 0} \frac{T(s_0)\times T(s_0+h)}h &= T(s_0) \times \lim_{h\to 0}\frac{T(s_0+h)-T(s_0)}h \\ &= T(s_0)\times T'(s_0)= \kappa(s_0) T(s_0)\times N(s_0) \\ &= \kappa(s_0) B(s_0), \end{align*} as desired.