Finding the limiting distribution of $\frac{X_n -n}{\sqrt{2n}}$ [closed]

Let $X_n \sim \chi^2 (n)$ for all $n$. I would like to show when $n\to\infty$

$$\frac{X_n -n}{\sqrt{2n}}\to N(0,1)$$

where $N(0,1)$ is normal distribution. Could someone help me?

Let $\left(Z_i\right)_i$ be a sequence of i.i.d. random variable following standard normal distribution, such that $X_n = \sum_{i=1}^nZ_i^2$, then
$\frac{X_n-n}{\sqrt{2n}} = \frac{\left(\sum_{i=1}^nZ_i^2\right)-n}{\sqrt{2n}} = \frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2/n}} = \sqrt{n}\left[\frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2}}\right]$.
Since $n^{-1}\left(\sum_{i=1}^nZ_i^2\right)$ can be viewed as the sample mean of $Z_i^2$, which follows a $\chi^2(1)$, it follows that
$\frac{X_n-n}{\sqrt{2n}} =\sqrt{n}\left[\frac{n^{-1}\left(\sum_{i=1}^nZ_i^2\right)-1}{\sqrt{2}}\right] \overset{d}\to N(0,1)$.