# If $\lim_{t\to\infty}\gamma(t)=p$, then $p$ is a singularity of $\gamma$.

I'm trying to solve this question:

Let $X$ be a vectorial field of class $C^1$ in an open set $\Delta\subset \mathbb R^n$. Prove if $\gamma(t)$ is a trajectory of $X$ defined in a maximal interval $(\omega_-,\omega_+)$ with $\lim_{t\to\infty}\gamma(t)=p\in \Delta$, then $\omega_+=\infty$ and $p$ is a singularity of $X$.

The first part is easy because $\gamma$ is contained in a compact for a large $t$, my problem is with the second part, I need help.

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Just a guess ... Could you say that once you know that $\lim_{t\to\infty}\gamma(t)=p$ then $\gamma(t)-p=c_1\exp(-c_2t)$ with $c_1,c_2$ constants and $c_2>0$, and then, since $X$ at $\infty$ is the derivative of $\gamma(t)$ at $\infty$, then you have that $\dot \gamma(t)=-c_1\exp(-c_2t)$, which evaluated at $t=\infty$ gives the desired result??? –  user58533 May 15 '13 at 6:17
For $n \geq 0$, let $t_n \in (n,n+1)$ such that $\gamma'(t_n)=\gamma(n+1)-\gamma(n)$ (use mean value theorem).
Then $\gamma'(t_n)=X(\gamma(t_n)) \underset{n \to + \infty}{\longrightarrow} X(p)$ and $\gamma'(t_n) \underset{n \to + \infty}{\longrightarrow} p-p=0$. Thus $X(p)=0$.