Limit of independent sequence of Normal Distribution is Chi Distribution

Question

I am looking for some help to solve the following central limit theory question below

Let $X_1, X_2, \ldots$ be i.i.d. with $X_i ≥ 0$, $EX_i = 1$, and $\operatorname{var} (X_i) = \sigma^2 \in (0, \infty)$. Show that $2\left(\sqrt{S_n} − \sqrt{n}\right) \Rightarrow \sigma \chi $.

Answer 1

You know that $$ \sqrt{n}\left(\frac {S_n}n - 1\right) \to G\sim N(0,\sigma^2) $$Then via delta method:

$$2\left(\sqrt{S_n} - \sqrt{n}\right)= 2\sqrt{n}\left(\sqrt{\frac {S_n}n} - 1\right) \to G'\sim N(0,\sigma^2) $$