Let $\dot{\bf{x}}=f(\bf{x})$ be a dynamical system with $f: \mathbb{R}^n\rightarrow \mathbb{R}^n$ being continuous and locally Lipschitz. Suppose that a particular solution trajectory satisfies $\lim_{t\rightarrow \infty}\bf{x}$ $(t)=\bf{x^*}$. Show that $\bf{x^*}$ is an equilibrium point.
Solution so far
Observe $\lim_{t \rightarrow \infty} \bf{x}$ $(t)= \lim_{t \rightarrow \infty}f(\bf{x}$$(t-1))$. But $f$ is a contraction map and continuous. That is $ \bf{x^*} $$= f(\lim_{t \rightarrow \infty}\bf{x}$$(t-1))=f$$(\bf{x^*}$)
If $f(x^*)\ne0$, you can use the flow box theorem to find coordinates such that $f$ is constant in some neighborhood $U$ of $x^*$. Since $x(t)\to x^*$, we have $x(t)\in U$ for all sufficiently large $t$. On the other hand, on that neighborhood (in the new coordinates) all travels with constant velocity and so after some time $x(t)$ will leave $U$. This contradiction shows that $f(x^*)=0$ and so $x^*$ must be an equilibrium.
Alternative without the flow box theorem:
If $f(x^*)\ne0$, then $c=\inf_{x\in U}\|Pf(x)\|>0$ in some neighborhood $U$ of $x^*$, where $P$ is the orthogonal projection on $f(x^*)$. Since $x(t)\to x^*$, we have $x(t)\in U$ for all sufficiently large $t$. On the other hand, on that neighborhood all travels with speed at least $c$ in the direction of $f(x^*)$ and so after some time $x(t)$ will leave $U$. This contradiction shows that $f(x^*)=0$ and so $x^*$ must be an equilibrium.