Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? Let $x'=f(t,x)$ be a differential equation with $f$ in the hypothesis of Picard's theorem. Let $\varphi$ be a solution such that its interval of definition contains $(t_0,+\infty)$ for some fixed $t_0\in \mathbb{R}$. Suppose $\lim_{t\to+\infty} \varphi(t)=a \in \mathbb{R}$. Must the constant function $t\mapsto a$ be a solution of the equation?
(This is a question I've asked myself while studying the logistic model $x'=ax(1-x)$ and wondering how to justify the solutions look the way they do.)
EDIT: By "the hypothesis of Picard's theorem", I mean $f$ continuous and locally Lipschitz with respect to $x$.
I'm interested also in particular cases (e.g. autonomous system) where this holds.
 A: You have not been explicit about "the hypotheses of Picard's theorem" (and yes, different sources state these differently) but in any case I believe that the answer is no unless you are far more restrictive on what you require of $f$.
Consider $f$ defined on $(1, \infty) \times \mathbb{R}$ by
$$
f(t,x) = \frac{\cos(t) - x}{t}, \qquad t > 1, \quad x \in \mathbb{R}.
$$
Whatever your hypotheses are I expect they are satisfied by this $f$ on $(1, \infty) \times \mathbb{R}$.
Consider $y: (1, \infty) \to \mathbb{R}$ defined by
$$
y(t) = \frac{\sin(t)}{t}, \qquad t > 1.
$$
A short calculation (just the quotient rule) shows that 
$$
y'(t) = \frac{t \cos(t) - \sin(t)}{t^2} = \frac{\cos(t) - \frac{\sin(t)}{t}}{t} = \frac{\cos(t) - y(t)}{t} = f(t,y(t)), \qquad t > 1,
$$
and clearly
$$
\lim_{t \to \infty} y(t) = 0.
$$
Yet $f(t,0) = \frac{\cos t}{t}$ is not identically zero, so the constant function $0$ is not a solution to $x' = f(t,x)$.
A: This is an elaboration of my previuous comment, written at Bruno's request.
Consider the equation $x'=f(x)$ with $f\colon\mathbb{R}\to\mathbb{R}$ continuous and let $x\colon(t_0,\infty)\to\mathbb{R}$ be a solution such that $\lim_{t\to\infty}x(t)=a$. I claim that $f(a)=0$, and hence $z(t)\equiv a$ is a constant solution. The proof is by contradiction.
Suppose $f(a)\ne0$. Without loss of generality we may assume that $f(a)>0$. We have
$$
\lim_{t\to\infty}x'(t)=\lim_{t\to\infty}f(x(t))=f(a).
$$
Let $\delta=f(a)/2$. There exists $t_1\ge t_0$ such that $x'(t)\ge\delta$ if $t\ge t_1$. In particular,
$$
x(t)=x(t_1)+\int_{t_1}^tx'(t)\,dt\ge x(t_1)+(t-t_1)\delta\quad\forall t\ge t_1,
$$
which contradicts the fact that $\lim_{t\to\infty}x(t)=a$.
The proof carries over to autonomous systems of equations.
