Vtyjkyuk

In one of my recent answers, I claim the following:

Since $dot r > 0$ when $0 < r < 1,$ and $dot r < 0$ when $r>1$, we may conclude that any solution starting in $(x_0,y_0)$ with $x_0^2+y_0^2 > 0$ will be attracted to the unit circle, so that
$$lim_t to infty r(t) = 1.$$

To briefly put this into context, we are working in polar coordinates, $r:mathbbRto [0,infty)$ is the radius function defined by $r(t)^2 = x(t)^2+y(t)^2$ and can be assumed to be smooth for the purposes of this question.

Of course, the result is intuitively clear, but I was unable to formulate a rigorous proof of this claim.
The thing that confuses me here is that we know the behaviour of the derivative depending on the value of $r$, rather than on the parameter $t$, so I am not sure how to use the information about the derivative in connection to the limit.

Could someone give me a hint on where/how to start?

EDIT: Just to be fully clear, I would like to prove the following:

Suppose that $r:mathbbRto(0,infty)$ is smooth and satisfies
beginalign
(1) quad dot r > 0 quad &textfor quad r < 1\
(2) quad dot r < 0 quad &textfor quad r > 1.
endalign
Then
$ lim_tto infty r(t) = 1. $

EDIT 2:
As pointed out in one of the answers, the claim is not true as stated.

This is wrong as it stands. Consider $r(t)=frac110arctan t$. Then $r(t)<1$ for all $t$ and $dot r(t)>0$ for all $t$ but $r(t)not to 1$ as $ttoinfty$.

You will need $r$ to satisfy some other conditions, for example a suitable ODE.

One way to approach this problem might be the following: If $r(t) = alpha in (0, 1)$, say, you may compute the derivative at this point. It will be an $epsilon$ away from $0$. You want to prove that for a sufficiently large interval, it remains away from $0$, so that the solution increases (by the differential equation, $r$ will stay below $1$ all the time, otherwise the derivative becomes positive).

So write down the equation
$$
r(1-r^2) le epsilon/2.
$$
What condition must $r$ satisfy so that this holds?

EDIT: For the general case, see my comment below.