Let $\alpha:I\to\mathbb{R}^3$ be an arc-length parametrized smooth space curve that is non-degenerate (i.e. $\alpha''\neq 0$). Fix a point $s_0\in I$. It is well-known that the osculating plane of $\alpha$ at $s_0$ is the limit of the plane containing three distinct points $\alpha(s_i)$, $i=1,2,3$ as $s_1,s_2,s_3\to s_0$. To state my question I shall review the argument.
Without loss of generality, assume $s_1<s_2<s_3$.
Let the plane determined by these three points be given by the equation \begin{align} \vec{n}\cdot\vec{x}-d=0 \end{align} where $\vec{n}=(a,b,c)$ is a normal to the plane, while $d\in\mathbb{R}$ is a constant. Note that both $\vec{n}$ and $d$ depend on $s_1,s_2,s_3$.
Now the key of the argument is to consider the function $f:I\to\mathbb{R}$ given by \begin{align} f(s)=\vec{n}\cdot\alpha(s)-d \end{align} Clearly $f$ is smooth. The assumption that $\alpha(s_i)$, $i=1,2,3$ lie on the plane means that \begin{align} f(s_1)=f(s_2)=f(s_3)=0 \end{align} So we can apply the Rolle's theorem in single variable calculus to conclude that \begin{align} \vec{n}\cdot\alpha'(\xi_i)=f'(\xi_i)=0,\qquad i=1,2 \end{align} for some $\xi_1\in(s_1,s_2)$ and $\xi_2\in(s_2,s_3)$. Then apply the Rolle's theorem again to conclude that \begin{align} \vec{n}\cdot\alpha''(\eta)=f''(\eta)=0 \end{align} for some $\eta\in(\xi_1,\xi_2)$. Now take limit as $s_1,s_2,s_3\to s_0$ (so $\xi_1,\xi_2,\eta\to s_0$ as well) to conclude that the plane is parallel to the vectors $\alpha'(s_0)$ and $\alpha''(s_0)$, and contains the point $\alpha(s_0)$, hence it is the osculating plane of $\alpha$ at $s_0$.
Now let us take a closer look on the limit-taking process, which leads to my question. When we prove that the limit plane is parallel to, says, the tangent vector $\alpha'(s_0)$, we are really proving that the normal of the limit plane is normal to $\alpha'(s_0)$, which is done by taking the limit of both sides of \begin{align} \vec{n}\cdot\alpha'(\xi_i)=0 \end{align} However, recall that the normal $\vec{n}$ is actually dependent of the points $s_1,s_2,s_3$, so when we take limit, besides having $\alpha'(\xi_i)\to\alpha'(s_0)$, we are also taking limit of $\vec{n}=\vec{n}(s_1,s_2,s_3)$. My question is:
How do we justify that this limit $\lim_{s_1,s_2,s_3\to s_0}\vec{n}(s_1,s_2,s_3)$ exists?
Or equivalently, how do we justify that the plane really converges to some limiting plane?
One may consider choosing $\vec{n}(s_1,s_2,s_3)$ that can be explicitly expressed in terms of $s_1,s_2,s_3$ and compute the limit of the expression. One natural candidate is the cross product $(\alpha(s_2)-\alpha(s_1))\times(\alpha(s_3)-\alpha(s_1))$, but clearly this tends to the zero vector when $s_1,s_2,s_3\to s_0$, which doesn't help the problem. One may then consider instead the normalized cross product \begin{align} \frac{(\alpha(s_2)-\alpha(s_1))\times(\alpha(s_3)-\alpha(s_1))} {|(\alpha(s_2)-\alpha(s_1))\times(\alpha(s_3)-\alpha(s_1))|} \end{align} which always has unit length, but it's not clear to me how to justify that this expression has a limit when $s_1,s_2,s_3\to s_0$.
Since this result is so standard in classical differential geometry of curves, I wonder if I'm missing something that's really simple. Any hint, comment or answer is welcomed and greatly appreciated.
I'm sure the two definitions are equivalent if $\alpha''$ is continuous in a nighbourhood of $s_0$. One should verify of course.
Coming back to your main question, you can even forget my comments because from $f'(\xi_1)=0$ and $f''(\eta)=0$ you get directly (!) $\,\vec n=\alpha'(\xi_1) \times \alpha''(\eta)\,$ up to a constant factor so that the limit is $\alpha'(s_0) \times \alpha''(s_0)$ by continuity.
Note that, if $\,\alpha'(s_0) \times \alpha''(s_0) \neq 0$, then $\alpha'\times \alpha'' \neq 0$ in a neighbourhood of $s_0$ by continuity.