Weak tangent but not a strong tangent Question: Show that $\alpha(t)=(t^3,t^2)$, $t\in \Bbb R$, has a weak tangent but not a strong tangent at $t=0$.
Definitions from this answer:
(Weak tangent) $\alpha: I \to \Bbb R^3$ has a weak tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0)$ has a limit position when $h \to 0$.
(Strong tangent) $\alpha: I \to \Bbb R^3$ has a strong tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0 + k)$ has a limit position when $h \to 0$ and $k \to 0$.
My query:
I'm not really clear what argument to use to demonstrate this. The weak tangent is the line joining $\alpha(t_0)$ and $\alpha(t_0+h)$, which is
$$
(\lambda(x(t_0+h)-x(t_0))+x(t_0), \lambda(y(t_0+h)-y(t_0))+y(t_0))
$$
If $t_0=0$ then $x(t_0)=0$ so this becomes
$$
(\lambda h^3, \lambda h^2)
$$
The strong tangent is
$$
(\lambda(x(t_0+h)-x(t_0+k))+x(t_0+k), \lambda(y(t_0+h)-y(t_0+k))+y(t_0+k))
$$
$$
=(\lambda (h^3-k^3)+k^3,\lambda (h^2-k^2)+k^2)
$$
As $h,k\rightarrow0$ this seems badly defined. But how can I make this argument precise?
Also, what is the intuitive meaning of the strong and weak tangents?
[This is exercise 1-3-7 of Differential Geometry of Curves and Surfaces by Do Carmo.]
 A: Weak tangent
Notice that the slope of $(\lambda h^3, \lambda h^2)$ tends to $\infty$ as $h\to 0$, meaning that the line direction approaches vertical. Since the line always passes through $(0,0)$, this means it has a limiting position (the $y$ axis).
Strong tangent
If it exists, it has to be the same as the weak tangent, because if the double limit exists, iterated limit "$k\to 0$ then $h\to 0$" exists and is equal to it. However, approaching via $h=-k$ you will find that the lines stay horizontal.
Intuitive meaning

*

*Strong tangent: if you walk along  the curve and someone is walking along the tangent line with the same speed, you can spend some  time walking together and holding hands.


*Weak tangent: looks like strong tangent at first, but at the point of tangency there is a break-up and someone goes away in the opposite direction.
A: Maybe one comment is worth here. Observe that the curve is $y = x^{2/3}$ and $dy/dx = (2/3)x^{-1/3}$ goes to $\infty$  when $x \rightarrow 0$, so the derivative does not exist at $x=0$ and $\alpha(t)$ is not regular at $t=0$. The same result should be obtained by the definition of weak and strong tangent if they exist. As mentioned before, the slope of the line determined by $\alpha(t_0+h)$ and $\alpha(t_0)$ goes to $\infty$ when $h\rightarrow 0$. So the weak tangent exists. But this does not happen for any $h$ and $k$ in the case of the strong tangent (the slope is $0$ if you choose $h=-k$).
