Show that Stabilizer of Lie group action is closed lie subgroup

Question

Let $G$ be a Lie group and $M$ a smooth manifold on which $G$ acts. For $x \in M$, how do I show that $G_{x} = \{ g \in G : g \cdot x = x\}$ is a closed lie subgroup? What I was thinking is that $f : G \times M \rightarrow M$ defined by $(g,m) \mapsto g \cdot m$ is a continuous map. Let $ \{ g_{n} \}_{n \in \mathbb{N}} \subset G_{x}$. By continuity of $f$, we have that $f(\lim_{n} g_{n},x) = \lim_{n \rightarrow \infty} f(g_{n},x) = \lim_{n} x = x$, which implies that $\lim g_{n} \in G_{x}$, so $G_{x}$ contains all of its limit points and hence is closed. Is this reasoning correct?

Answer 1

To show that $G_x$ is closed, you need to show that if the limit of $g_n$ is $g$, $g_n\in G_x$, then $g\in G_x$, $lim_nf(g_n,x)=lim_nx=x=f(g,x)$ implies $g\in G_x$.

You can also remark that $f_x:G\rightarrow M$ defined by $f_x(g)=f(g,x)$ is continuous and $G_x=f^{-1}(x)$ is closed since $\{x\}$ is closed.