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 $t \to \infty$?

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.
