How does smoothness prevent "singularities"? This is a refinement of one of my earlier questions (I failed to put into words what I really wanted to ask).
First of all, I'm not sure "singularity" is the correct word to use hence the quotes. Consider the following wild knot:

Then what exactly happens where the curls get infinitely small? Is the knot still differentiable there? I'm asking because I'm trying to understand why requiring a knot to be differentiable is not enough to prevent knots from being wild. On the other hand, smoothness is enough. 
Thanks for help!
(If anyone knows the parametric equation of this curve it might make it easier to see what happens.)  

This is not really an answer so much as me writing out some confusion, so I'm making it community wiki. Consider the curve
$$
f(x) = \begin{cases} x^2\sin(1/x) & x < 0 \\
0 & x \geq 0.
\end{cases}
$$
Then $f$ exhibits wild-knot like behavior at the origin and is differentiable but not twice differentiable there; you could probably make your wild knot parametrically by bending the graph of $f$ into three-space. But on the other hand 
$$
f(x) = \begin{cases} \exp\left(\frac{-1}{x^2}\right)\sin(1/x) & x < 0 \\
0 & x \geq 0.
\end{cases}
$$
exhibits infinite oscillation and is smooth, so it seems like I ought to be able to make a smooth wild knot too, if my argument in the earlier part is right.
