Consider the following nonliner system: \begin{align} \dot{x}=f(x) \end{align} where $x\in\mathbb{R}^n$ and $f(x)\in\mathbb{R}^n$ is sufficiently smooth and Lipschitz in $x$. Then the system is smooth and admits a unique solution. Suppose $x^*\in\mathbb{R}^n$ is an equilibrium of the system, i.e., $f(x^*)=0$. Is it possible that, for some initial condition $x(0)=x_0$, the solution of the system satisfies \begin{align} \lim_{t\to T}x(t)=x^{*}, \end{align} that is, the solution reaches the equilibrium $x^*$ in some finite time $T$. If it is not possible, is there a way to show that, the solution of a smooth system will take an infinite amount of time to converge to an equilibrium?
Update：Assume that $x_0\neq x^{*}$.
Let $y(t) = x^*$ and note that $y$ is a solution to the system $\dot{y} = -f(y)$ with $y(0) = x^*$.
Suppose $x$ is a solution with $x(T) = x^*$ and $x(0) = x_0$. Then since $x(T)=y(T)$ we must have $x(t) = y(t)$ for $t <T$ (by uniqueness, running the system backwards, that is $\dot{z} = -f(z)$). Hence $x_0 = x^*$.
In particular, if $x_0$ is not an equilibrium and $x(t) \to y^*$ where $y^*$ is an equilibrium, then $x(t) \neq y^*$ for all $t$.