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.
Thanks in advance

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 $X(p)=0$.
