The vector $\mathbf{r}'(t)$ is called the tangent vector to the curve defined by $\mathbf{r}$ at the point $\mathbf{r}(t)$, provided that $\mathbf{r}'(t)$ exists and $\mathbf{r}'(t)\neq\mathbf{0}$.
Let $\mathbf{r}(t):=(t^3,t^7)$.
Then, $\mathbf{r}'(0)=\mathbf{0}$.
So, the tangent vector to the curve defined by $\mathbf{r}$ at the point $\mathbf{r}(0)$ is not defined.
$\mathbf{r}'(t)=(3t^2,7t^6)$.
Suppose $\epsilon$ is a nonzero real number whose absolute value is very small.
$\mathbf{r}'(\epsilon)=(3\epsilon^2,7\epsilon^6)=3\epsilon^2(1,\frac{7}{3}\epsilon^4)\approx 3\epsilon^2(1,0)$ since $\epsilon^2\ll\epsilon^4$.
So, I want to say $(1,0)$ is the unit tangent vector to the curve defined by $\mathbf{r}$ at the point $\mathbf{r}(0)$.
Why is my idea bad?