If a smooth integral curve has a limit point $p$ and it takes infinite time to reach that point. Does that imply that its velocity approaches $0$ at that point?
Or in other words,
Updated based on comment suggestions
Given a smooth vector field $V: \mathbb{R}^n \rightarrow \mathbb{R}^n$. And an integral curve $c(t)$ satisfying.
$$c'(t) = V(c(t))$$
Let $p$ be a point in $\mathbb{R}^n$.
Then, $$\lim_{t \rightarrow \inf} c(t) = p \implies \lim_{t \rightarrow \inf} c'(t) = 0$$
In otherwords, $p$ must be a singular point of the vector field $V$.
I've struggled to prove this as every proof I have seen requires that the derivative be bounded or uniformly continuous, I believe. We know that $c$ is locally lipschitz and smooth. Is there a proof that follows from those two conditions?
Update I know that the above can be proved by using the machinery of flows, but I was trying to avoid that.
Edit We can also assume that $\lim_{t\rightarrow \infty} c'(t)$ exists, since this is an integral curve of a smooth vector field and it approaches a point on $\mathbb{R}^n$.
The point sequence $c(n)$ converges to $c^*$ due to the asymptotic convergence of the function to $c^*$.
$c$ is continuously differentiable. Thus there are middle points $\theta_n\in(n,n+1)$ with $c'(θ_n)=c(n+1)-c(n)$ via the mean value theorem.
Per the first point, $c'(θ_n)$ converges to zero.
$c(θ_n)$ also converges to $c^*$.
Due to the continuity of $V$, $V(c^*)=\lim_{n\to\infty}V(c(θ_n))=\lim_{n\to\infty}c'(θ_n)=0$.
Again by the continuity of $V$, $\lim_{t\to\infty}c'(t)=\lim_{t\to\infty}V(c(t))=V(c^*)=0$.