Let $\dot{x} = f(x)$ where $x \in \mathbb{R}^n$ be a dynamical system with $f : E \rightarrow \mathbb{R}^n$ being continuous on the open set $E$. Suppose a particular trajectory $x(t)$ satisfies $\lim_{t \rightarrow\infty}x(t) = \bar{x}$. Prove that $\bar{x}$ is an equilibrium.
My try:
We need to show $\dot{x}(t) = f(x(t))$ as $t \rightarrow \infty$ is zero. So we just need to show $\dot{x}(t) = 0$ as $t \rightarrow \infty$
$$ \dot{x}(t) = \lim_{h \rightarrow 0} \frac{x(t+h) - x(t)}{h} = 0 \,\,\,\,\,\,,\text{as} \,\,\,\,\,\,t \rightarrow \infty $$
which intuitively make sense because $\lim_{t \rightarrow\infty}x(t) = \bar{x}$. However, I do not know how to use $\lim_{t \rightarrow\infty}x(t) = \bar{x}$ and apply it on the limit. We can say since $\lim_{t \rightarrow\infty}x(t) = \bar{x}$, for all $\epsilon >0$ there exist a $N \in \mathbb{N}$ such that for $t \geq N$ $$ \|x(t) - \bar{x}\| \leq \epsilon $$
How can we use the above to have things rigorously.