Suppose an ODE's system:
$\begin{cases}dx/dt=f(x,y,z,t)\\ dy/dt=g(x,y,z,t)\\ dz/dt=h(x,y,z,t)\\ (x(0),y(0),z(0))=(a,b,c)\end{cases}$
If I am asked to find the equilibrium points, I'll try solve:
$\begin{cases}0=f(x,y,z,t)\\ 0=g(x,y,z,t)\\ 0=h(x,y,z,t)\end{cases}$
But I've read at some texts that people say:
For an equilibrium state $(x^*,y^*,z^*)$ to exist, the following limits must exist: $$\lim_{t\to\infty} x(t)=x^*;\lim_{t\to\infty} y(t)=y^*;\lim_{t\to\infty} z(t)=z^*.$$
I'd like to know the differences and similarities between these two thoughts.
In particular, in some cases the limit exists, but equality does not.
Many thanks.
While I was writing up my answer, I realized that my comment on the OP missed the actual difference between the two situations. Let me switch to a more general (and easier to deal with) notation than you have used: Given a map $F : \Bbb R^d\times [0,\infty) \to \Bbb R^d$, we are looking for curves $v : [0,\infty) \to \Bbb R^d$ which satisfy $$\frac {dv}{dt} = F(v, t)\\v(0) = v_0 \in \Bbb R^d$$
In general the idea of equilibrium is a point that a solution $v$ will want to be at. Either it will stay at that point, or will approach it as time goes by. This is what the limit definition is getting at: $v^*$ is an equilibrium point if for some solution $v, \lim_{t\to \infty} v(t) = v^*$.
Any point $v^*$ for which $F(v^*, t) = 0$ for all $t \in [0,\infty)$ will be an equilibrium state, since $v(t) = v^*$ is a solution to the differential equation, and obviously, $\lim_{t \to \infty} v^* = v^*$.
Some of those equilibrium points will be stable: if $v_0$ is anywhere near $v^*$, then $\lim_{t \to \infty} v(t) = v^*$. Some will be unstable: for any $v_0 \ne v^*, \lim_{t \to \infty} v(t) \ne v^*$. This does not change the fact that at $v_0 = v^*$, the limit is $v^*$, so the point is still an equlibrium point. And others will be semi-stable: for some $v_0$ near $v^*$ the limit is $v^*$, and for other $v_0$ near $v^*$ it is not.
But can there exist equilibrium points $v^*$ for which $F(v^*, t) \ne 0$ for an unbounded collection of $t \in [0, \infty)$? Yes. Here is a one-dimensional example: $$F(x,t) = -x + e^{-t}\sin t$$ You can easily check that $v(t) = e^{-t}(v_0 - \cos t)$ is a solution. $\lim_{t \to \infty} v(t) = 0$. But $F'(0, t) = 0$ only when $t = \frac \pi2 + k\pi, k \in \Bbb N$.
The $-x$ term in $F$ pushes $v$ towards $0$, but the $e^{-t}\sin t$ term keeps perturbing it, so it can't settle down. But since that perturbance decreases with time, in the limit $v$ is finally able to converge.
So, $\lim_{t\to\infty} v(t) = v^*$ for some $v_0$ is what defines equilibrium points. Solving $F(v^*, t) = 0$ is a way of finding equilibrium points, but is not guaranteed to find them all when $F$ depends on $t$.
(If $F$ is well enough behaved and does not depend on $t$, then I think you can show that $F(v^*) = 0$ will give all equilibrium points.)