Supose that $S_n$ has a $\chi^2$ distribution with $n$ degrees of freedom. Show that $ \mathbb{P}(S_n \le x) = f\left(\sqrt{2x}-\sqrt{2n}\right) $ where $f(u)$ is the normal distribution.
I tried this: $\mathbb{P}(S_n \le x)= \mathbb{P}(\sqrt{2 S_n}-\sqrt{2 n} <= \sqrt{2x}-\sqrt{2n})$ now I am trying to show that $\sqrt{2 S_n} - \sqrt{2n}$ converges in law to $\mathcal{N}(0,1)$.