Consider the curve given by
$\alpha(t) = \left\{ \begin{array}{ll} (t,0,e^{-1/t^2}) & t > 0 \\ (t,e^{-1/t^2},0) & t <0 \\ (0,0,0) & t = 0 \end{array} \right. $
Problem: Prove that $\alpha$ is regular for all $t$.
Since a curve is regular at $t_0$ iff $\alpha'(t_0) \neq 0$ the curve seems to violate this at $t=0$. Perhaps this has something to do with the curve not being continuously differentiable at that point, but I have a feeling I'm missing something big here.
Well, $\alpha$ is regular for $t \neq 0$ since the first component of $\alpha'(t)$ is $1\neq 0$. But this actually holds for $t=0$ too (use the definition of derivative in case you don't see it immediately).