# Strong/weak tangents and limit positions, with rigor

As I'm working from do Carmo's Differential Geometry of Curves and Surfaces, I have found some of his imprecise language regarding strong and weak tangents to be most irksome. I've seen similar posts floating around this site from those suffering the same woe. See my comments here.

In do Carmo's words:

Let $$\alpha : I \to \mathbb{R}^3$$ be a simple curve of class $$C^0$$ (i.e. continuous) ... We say that $$\alpha$$ has a strong tangent at $$t=t_0$$ if the line determined by $$\alpha(t_0+h)$$ and $$\alpha(t_0+k)$$ has a limit position when $$h,k \to 0$$.

(Note: we'd obviously like to readily generalize to $$\mathbb{R}^n$$.)

Nowhere in the text does do Carmo actually define a "limit position," but it seems easy to deduce what he's going for.

Is the following a valid, rigorous definition of a strong tangent?

We say that $$\alpha$$ has a strong tangent at $$t=t_0$$ if the limit $$\lim_{(h,k) \to \vec{0}} \dfrac{[\alpha(t_0+h)-\alpha(t_0+k)]/(h-k)}{\left\lvert [\alpha(t_0+h)-\alpha(t_0+k)]/(h-k) \right\rvert}$$ exists and is not equal to the zero vector (such that the limit determines a line).

I came to that definition after playing around with limits in Mathematica for a good while. If that doesn't work, then what does? Is the normalization inside the limit necessary?

A motivating example is the curve $$\alpha(t)=(t^3,t^2)$$ with $$t \in \mathbb{R}$$, which has a weak tangent at $$t=0$$, but not a strong tangent. It's easy to examine the slope $$[y(t_0+h)-y(t_0+k)]/[x(t_0+h)-x(t_0+k)]$$, but that method doesn't easily generalize to $$\mathbb{R}^n$$.

• After working in Mathematica some more, it seems that even the above definition does not suffice, because if we define a weak tangent in a similar fashion, we find that $\alpha$ in the example above does not have a weak tangent at $t=0$ (when in fact it should/does). – terrygarcia Mar 13 at 21:06
• One can take those words about the "limit position" of a line quite literally, using that the set of lines in $\mathbb R^3$ has a very natural topology, in fact it has a natural structure as a smooth manifold of dimension $4$. One can work out local coordinates, and then just take a limit in those coordinates. – Lee Mosher Mar 13 at 21:10
• @LeeMosher It seems that you're getting at the real projective space $\mathbb{R}P^5$ formed from $\mathbb{R}^3 \times \mathbb{R}^3 \simeq \mathbb{R}^6$. I'm not too familiar with projective geometry; could you provide an example of your method, possibly as an answer? – terrygarcia Mar 13 at 21:19
• Never mind that, I see you edited to "of dimension 4" – terrygarcia Mar 13 at 21:20