Suppose I have a function $P:[0, \infty)\to K$ where $K$ is a compact set (not necessarily a subset of $\mathbb R$). Suppose that I know (for other reasons) that $P(t)$ has a unique possible limit $c\in K$. This means that if it exists $\lim_{t\to\infty}P(t)$ then $\lim_{t\to\infty}P(t)=c$. Can I conclude that $$\lim_{t\to\infty}P(t)=c?$$
My doubt comes from the fact that if I had a sequence $P_n\in K$ and I knew that there is a unique possible limit in $K$ then I could say that the sequence $P_n$ is tight (because $K$ is compact),and that it converges by subsequences. But every subsequence converges to the same limit $c$ and therefore I can conclude that $$\lim_{n\to\infty}P_n=c.$$ There is something similiar that holds for the function $P(t)$?
@edit
I can change my question in the following way.
Suppose I have a trajectoty $(P_t)_{t\in [0, \infty)}$. However I take a sequence $(P_{t_n})_{n\in \mathbb N}$ such that $\lim_{n\to\infty}t_n=\infty$ then $\lim_{n\to\infty}P_{t_n}=c$. Can I conclude that $$\lim_{t\to\infty}P_t=c?$$
I am not sure whether or not this answers your question, But I'll give it a try.
Take $\sin\colon[0,\infty)\longrightarrow[-1,1]$. Since $(\forall n\in\Bbb N):\sin(\pi n)=0$, then, if the limit $\lim_{x\to\infty}\sin(x)$ exits, it can only be $0$. But that limit does not exist.