Proving that an autonomous differential equation has a constant solution under a certain condition Let $f: \Omega \to \mathbb{R}^n$ be a continuous vector field, with $\Omega \subseteq \mathbb{R}^n$ open. Let $x: (a, ∞) \to \Omega$ be a solution of the autonomous differential equation $\dot{x} = f(x)$ that converges against an $x_0 \in \Omega$.
I'm supposed to show that $f(x_0) = 0$, meaning $x_0$ is a constant solution of the system, by integrating $\dot{x}(t)$ for large $t$.
I think I might have solved this problem, but my solution seems a bit too simple, and strangely, the "integrate $\dot{x}$"-part that is explicitly stated in the task seems a bit arbitrary to me if my solution is indeed right. Therefore, I wonder if I have done something wrong, or if this part in the assignment is indeed unnecessary; and I'd be glad if someone could take a look.
Now first, since $lim_{t \to \infty} x(t) = 0$, we have that each coordinate of $x(t)$ converges against the according coordinate of $x_0$, so proving that each coordinate of $f(x_0)$ is $0$ would give me the required result.
To show that, I first followed the "hint" to integrate $\dot{x}(t) = f(x(t))$. It seems that the arising integral would then approach $0$ as much as we want: for example, we would have that $lim_{t \to \infty} \left\lvert\int_t^{t+1} f_i(x(s))ds\right\lvert = 0$. Hence, any antiderivative of $f_i$ would very much approach a constant function. Now it seems to me that this would already be sufficient to argue that, because $f$ (and therefore each $f_i$) is continuous, we have that $f_i(x_0) = 0$ because $lim_{t \to \infty} f_i(x(t)) = 0$. That seemed plausible to me.
But wait. Couldn't we then drop the "integrate $\dot{x}(t)$"-part? If $lim_{t \to \infty} x_i(t) = c_i$ (with $c_i$ being the $i$-th coordinate of $x_0$, then $lim_{t \to \infty} \dot{x}_i(t) = 0 = lim_{t \to \infty} f_i(x(t))$, which would give the same result? Or am I overlooking something here? It seems strange to me to state this in the task if it is unnecessary in the end.
 A: We have $x(t) = x(t_0) + \int_{t_0}^t f(x(\tau)) d \tau$, and $x(t) \to x_0$.
Since $f$ is continuous, we have $f(x(t)) \to f(x_0)$.
Suppose $f(x_0) \neq 0$. Let $\phi(y) = f(x_0)^T y$, we note that $\phi(f(x_0)) > 0$. By continuity, there is some $\epsilon>0$ and $T$ such that if $t \ge T$
then $\phi(f(x(t))) \ge \epsilon$.
Now apply $\phi$ to the integral equation above, using $t_0 = T$ to get
$\phi(x(t)) = \phi(x(T)) + \int_T^t \phi(f(\tau)) d \tau \ge \phi(x(T)) + (t-T) \epsilon$, which contradicts the
fact that $\phi(x(t)) \to \phi(x_0)$.
A: As the OP asks for a feedback on the solution, I would like to comment on two passages that I have found incorrect.

  
*
  
*$\lim_{t \to \infty} \left\lvert\int_t^{t+1} f_i(s)ds\right\lvert = 0^{*)}$. Hence, any antiderivative of $f_i$ would very much approach a constant function.
  

The counterexample: $f_i(s)=\frac{1}{s}$ with an antiderivative $F_i(s)=\ln s$. We have
$$
\int_t^{t+1}\frac{1}{s}\,ds=\ln(t+1)-\ln (t)=\ln\frac{t+1}{t}\to 0\quad\text{as}\ t\to\infty,
$$
but the antiderivative does not approach a constant value.


  
*If $\lim_{t\to\infty}x_i(t)=c_i$ [...] then $\lim_{t\to\infty}\dot x_i(t)=0$.
  

The counterexample: take 
$$
x_i(t)=\frac{\sin(t^2)}{t}\to 0,\qquad\dot x_i(t)=\frac{\cos(t^2)2t^2-\sin(t^2)}{t^2}=2\cos(t^2)-\frac{\sin(t^2)}{t^2}\not\to 0.
$$

*) Shouldn't here be $f_i(x(s))$ in the integral?
