Closure of $\{ ( e^t\cos t,e^t\sin t) : t \in \Bbb R \}$

Question

Suppose $\alpha: \Bbb R\to \Bbb R^2$ given by $\alpha (t)=(e^t \cos t,e^t \sin t)$, $A=\alpha(t)$ is a smooth manifold. What is the closure of $A$?

I know that the closure of the set is the set itself and plus all its limit points. How can I apply the theorem here?

Answer 1

The only point that you get close to but not reach is $(0,0)$ which is the limit as $t \to - \infty$. So the closure is just the image plus this one point.