Extension of smooth maps at a cusp There is a short remark in deCarmos "Riemannian Geometry" (p. 67)

and I wonder about the condition that the vertex angles must be $\neq \pi$. If $s_1$ and $s_2$ are two differentiable maps on an open set containing $A$ and if $p$ is a cusp point in $\partial A$, then how is it possible that $ds_1|_p \neq ds_2|p$? 
Taking the first open set $U$ in the above quote, what about a sequence $x_n$ in $U$ sucht that $x_n \rightarrow p$ and then by continuity  $ds_1|_p = ds_2|p$? Where does the condition about the angle $\neq \pi$ enter the discussion?
 A: The question is whether the values of $s$ on $A$ determine $Ds_p$ for all $p \in A$. Knowing  $Ds_p$ is equivalent to knowing $Ds_p (v)$ for two linearly independent $v$. Knowing $Ds_p (v)$ is equivalent to knowing $s(p + t_n v)$ for some sequence $0<t_n \to 0$.
If $p \in U$ then there is no issue - we can take an open ball around $p$ contained in $A$. If $p\in \partial A$ is not a vertex, then exactly half the vectors $v \in T_p M$ have rays $p + t v$ that stay in $A$ for some positive time, and thus the values of $s$ on $A$ easily determine $Ds_p$.
When $p$ is a vertex with turning angle $\theta$, there is a cone of vectors with angle $\pi - \theta$ with rays staying in $A$ for a bit, so as long as $\theta < \pi$ there are two linearly independent such vectors.
The issue when $\theta = \pi$ is that the values of $s$ on $A$ only determine $Ds_p (v)$ for a single direction $v$ - in every other direction there is a short time for which the rays are disjoint from $A$, and thus the behaviour of $Ds_p$ in these directions is not determined by $s|_A$. I don't have time to work out a specific example but it shouldn't be too hard.
I believe your argument using sequences rules out this possibility if you assume that $s$ is continuously differentiable.
