Then find the limit of r(t)

This is an indeterminate form $\frac{0}{0}$. We can factor the numerator:

This function is continuous at $t=1$, so we can substitute the value:

As $t \to 1$, the numerator $\sin(\pi t) \to \sin(\pi) = 0$, and the denominator $\ln t \to \ln(1) = 0$. This is an indeterminate form $\frac{0}{0}$, so we can use L'Hôpital's Rule.
Take the derivative of the numerator and the denominator:
Derivative of numerator: $\frac{d}{dt}(\sin(\pi t)) = \pi \cos(\pi t)$
Derivative of denominator: $\frac{d}{dt}(\ln t) = \frac{1}{t}$
Now, apply L'Hôpital's Rule:

The limit of $\vec{r}(t)$ as $t \to 1$ is: