I would like to propose the following definition, which may not work in the most general scenario but seems to be sufficient for the purpose of discussing the weak and strong tangents.
Let $I\subseteq\mathbb{R}$ be an open interval, and let $h_0,k_0\in I$.
Let $\{\ell_h\}_{h\in I'}$ ($I'$ can possibly not containing $h_0$) be a 1-parameter family of straight lines, each of which is given by the equation
\begin{equation}
\mathbf{x}=\mathbf{p}(h)+t\cdot\vec{v}(h)
\end{equation}
Of course the same line $\ell_h$ may be represented by different choices of $\mathbf{p}(h)$ and $\vec{v}(h)$. We now say that $\{\ell_h\}_{h\in I}$ has a limiting position as $h\to h_0$ if:
- There exists a choice of $\{\mathbf{p}(h)\}_{h\in I'}$ such that the limit
\begin{align}
\mathbf{p}_0:=\lim_{h\to h_0}\mathbf{p}(h) & & (1.1)
\end{align}
exists. (Since each $\mathbf{p}(h)=(x(h),y(h),z(h))$ is a point in $\mathbb{R}^3$, it makes sense to talk about their limit.)
- The limit
\begin{align}
\vec{v}_0:=\lim_{h\to h_0}\frac{\vec{v}(h)}{|\vec{v}(h)|} & & (1.2)
\end{align}
exists. (The normalization is taken to prevent the limit diverge to infinity due to the increasing norm of $\vec{v}(h)$. After all, the only information we care about is the direction, and we would not like the increasing norm to disturb the extraction of this information.)
Then, we call the line $\mathbf{x}=\mathbf{p}_0+t\cdot\vec{v}_0$ the limiting line of the family $\{\ell_h\}_{h\in I'}$.
Under this proposed definition, we can check that the curve $\alpha(t)=(t^3,t^2)$ indeed has a weak tangent at $t=0$ as claimed in do Carmo.
Next, let $\{L_{h,k}\}_{h,k\in I'}$ be a $2$-parameter family of straight lines, each of which has similar equation
\begin{align}
\mathbf{x}=\mathbf{p}(h,k)+t\cdot\vec{v}(h,k)
\end{align}
The definition is similar: We say that $\{L_{h,k}\}_{h,k\in I'}$ has a limiting position as $h\to h_0$ and $k\to k_0$ if
- There exists a choice of $\{\mathbf{p}(h,k)\}_{h,k\in I'}$ such that the limit
\begin{align}
\mathbf{p}_0:=\lim_{(h,k)\to(h_0,k_0)}\mathbf{p}(h,k) & & (2.1)
\end{align}
exists; and
- the limit
\begin{align}
\vec{v}_0:=\lim_{(h,k)\to(h_0,k_0)}\frac{\vec{v}(h,k)}{|\vec{v}(h,k)|} & & (2.2)
\end{align}
exists,
in which case we again call the line $\mathbf{x}=\mathbf{p}_0+t\cdot\vec{v}_0$ the limiting line of the family $\{L_{h,k}\}_{h,k\in I'}$.
Then, under this proposed definition, the curve $\alpha$ above does not have a strong tangent at $t=0$. The problem, of course, is that the limit $\vec{v}_0$ does not exist, essentially due to the fact that the limit of a multivariable function is a strong notion, in the sense that a function failing the so-called "two-paths test" in multivariable calculus must not have a limit.
I have not spent much time to examine these proposed definition. It is perfectly possible to have some examples which intuitively should have a limiting position but fails to do so under my proposed definition (or the other way round -- intuitively not having the limiting position but survives under my proposed definition). Worse, I did not even check if the limiting line is unique: Could it be different if we take a different choice of $\{\mathbf{p}(h)\}_{h\in I'}$?
Feel free to let me know any of such examples, and let us use them to keep refining the definition.
EDIT: It turns out that the condition (2.2) is too strong $--$ The proposition that "Every simple curve of class $C^1$ that is regular at $t=t_0$ has a strong tangent at $t=t_0$" in do Carmo fails to hold under this proposed definition. To see this, consider the very trivial example $L_{h,k}$ passing through $(0,0,h)$ and $(0,0,k)$ (where $h\neq k$):
\begin{align}
\mathbf{x}=(0,0,h+t(k-h))
\end{align}
so that $\vec{v}(h,k)=(0,0,k-h)$. Along the two paths $\{h=0,k>0\}$ and $\{h>0,k=0\}$, the limits are
\begin{align}
\lim_{\underset{h=0,k>0}{(h,k)\to(0,0)}}\frac{\vec{v}(h,k)}{|\vec{v}(h,k)|}=(0,0,1)
\qquad\text{but}\qquad
\lim_{\underset{h>0,k=0}{(h,k)\to(0,0)}}\frac{\vec{v}(h,k)}{|\vec{v}(h,k)|}=(0,0,-1)
\end{align}
and so (2.2) fails to hold. However, every line in $\{L_{0,k}\}_{k>0}$ and $\{L_{h,0}\}_{h>0}$ is the $z$-axis and so they "converge" trivially to the $z$-axis as $k\to 0$ and $h\to 0$ respectively.
The problem lies in the fact that both $\vec{v}(h,k)$ and $-\vec{v}(h,k)$ give the same line, but the condition (2.2) does not take into account of this. As a result, the limit along different paths may not be unique. However, if a family of lines intuitively has a limiting line, then we should expect that the sign difference is the only ambiguity that could arise in the computation of limits.
In particular, if the limiting line has direction given by a unit vector $\vec{v}_0$, then as $h,k\to 0$, if $\vec{v}(h,k)/|\vec{v}(h,k)|$ is not close to $\vec{v_0}$, then we should expect it to be close to $-\vec{v}_0$. Equivalently, the inner product $\left\langle\frac{\vec{v}(h,k)}{|\vec{v}(h,k)|},\vec{v}_0\right\rangle$ should be close to either $1$ or $-1$.
In view of this, I would replace (2.2) by the following:
There exists a unit vector $\vec{v}_0$ such that
\begin{align}
\lim_{(h,k)\to(h_0,k_0)}\left|\left\langle\frac{\vec{v}(h,k)}{|\vec{v}(h,k)|},\vec{v}_0\right\rangle\right|=1 & & (2.2')
\end{align}
With this, it is now true that every simple curve of class $C^1$ that is regular at $t=t_0$ has a strong tangent at $t=t_0$.
