A strange $C^{\infty}$ curve Consider the function $f : \mathbb R\to\mathbb R$ defined as $f(x)=e^{-1/x^2}$ for $x>0$ and $0$ for $x\le0$. This is easily seen to be a $C^{\infty}$ function. Now,
define functions $f_1,f_2,g_1,g_2:\mathbb R\to\mathbb R$ as $f_1(x)=f(x-1)$ and $f_2(x)=f(-x-1)$ and $g_1=f_1-f_2$ and $g_2=f_1+f_2$. Finally, define the curve $g:\mathbb R\to\mathbb R^2$ by $g(x)=(g_1(x),g_2(x))$. The range of this curve is a subset of the graph of the function $h:\mathbb R\to\mathbb R$ given by $h(x)=|x|$. Intuitively, this curve should not have a well defined tangent at the point $(0,0)$. But, we have a smooth parameterization here. Does this mean that there is no way to define the tangent line to a curve, independent of parametrization?

But what do you get when you compute the derivative, at that point, of the parametrization? 
Does it (the derivative) define a direction?
Let us compute
$$\begin{align}g'(0)&=(g_1'(0),g_2'(0))\\&=(f_1'(0)-f_2'(0),f_1'(0)+f_2'(0))\\&=(f'(-1)+f'(-1),f'(-1)-f'(-1))\\&=(0,0)\end{align}$$
Does this vector define a direction of the tangent?
The moral is, for the tangent line to be defines the parametrization must be not only differentiable, but also give a non-zero derivative.
