How to estimate a limit value of ordinary differential equations $$\frac{dx}{dt}=(x-y)(1-x^2-y^2)\\
\frac{dy}{dt}=(x+y)(1-x^2-y^2)$$
Initial condition $x(0),y(0)$ are nonzero real numbers. 
How to estimate the solution $x(t)$ as $t \to \infty$?
 A: If $z=x+iy$ then
$$\dot z = (1+i)z(1-|z|^2)$$
so that
$$
\frac{d}{dt}\arg z = Im(\frac{\dot z}{z})=1-|z|^2=Re(\frac{\dot z}{z})=\frac{d}{dt}\ln|z|
$$
which integrates to
$$\arg(z(t))-\arg(z(0))=\ln|z(t)|-\ln|z(0)|.$$
If, as per the other answers, $|z(t)|\to 1$ for $t\to\infty$ then 
$$\arg(z(\infty))=\arg(z(0))-\ln|z(0)|,$$
all angles up to multiples of $2\pi$.
A: The common factor $(1-x^2-y^2)$ is just a rescaling of the vector field (with different signs inside and outside the unit circle). So the orbits agree, up to time parametrization and possibly a change of direction, with those of the linear system $dx/dt=x-y$, $dy/dt=x+y$. (And the behaviour on the unit circle is of course different too.) So if you can solve the linear system, then you should be able to figure out the answer.
Compare the phase portraits (on Wolfram Alpha): linear, rescaled.
A: Note that if $(x(0),y(0))= (0,0)$ or $x^2(0)+y^2(0) = 1$, then
$(x(t),y(t))=(x(0),y(0))$ for all $t$.
Let $s(t) = x^2(t)+y^2(t)$, then
$\dot{s} = f(s)=2(1-s)s$.
Note that $f(0)=f(1) = 0$, if $s \in [0,1]$ then $f(s) \ge0$ and
if $s \ge 1$ then $f(s)  \le 0$.
In particular, if $s(0) >0$ then $\lim_{t \to \infty} s(t) = 1$. In fact, since this is a one dimensional ODE, we know that if $s(0) \in (0,1]$ then $\lim_{t \uparrow \infty} s(t) = 1$ and if
$s(0) \ge 1$ then $\lim_{t \downarrow \infty} s(t) = 1$.
We can find an explicit estimate for the convergence of $s(t)$. Note that for $s\ge 1$ we have $f(s) \le 2(1-s)$ and so $\dot{(s-1}) \le -2 (s-1)$ which gives $0 \le s(t)-1 \le (s(t_0)-1) e^{-2(t-t_0)}$ for $s(t_0) \ge 1$.
Note that for any $\alpha <2$ we can find some $\delta>0$ such that if $s \in [1-\delta,1]$ then $f(s) \ge \alpha (1-s)$. Hence if $s(t_0) \in [1-\delta,1]$, we have $0 \le 1-s(t) \le (1-s(t_0)) e^{-\alpha(t-t_0)}$.
Hence for sufficiently large $t_0$ we have the estimate
$|1-s(t)| \le |1-s(t_0)| e^{-\alpha(t-t_0)}$.
If we let $w=(x,y)$ then we can write the ODE as
$\dot{w} = (1-\|w\|^2) Aw$ for some matrix $A$ and we are given $w(0) \neq 0$. Let $s(t)= \|w(t)\|^2$, then we note that $s(t) \to 1$ hence
the solution is bounded, and so there is some $K$ such that
$\|\dot{w}\| \le K \|A\| |1-s(t)|$, in particular $t \mapsto \|\dot{w}(t)\|$ is integrable and hence $w$ converges.
