According to Wikipedia, the (centered and scaled) chi-distribution converges to the standard normal distribution:
$$\frac{\chi_k-\mu_k}{\sigma_k} \xrightarrow{d} \mathcal N(0,1)$$
Where by definition $$\chi_k = \sqrt{X_1^2+ \ldots + X_k^2}$$ where $X_j\sim\mathcal N(0,1)$ are i.i.d.
The mean and variance (also from Wikipedia) are:
\begin{align} \mu_k & = \sqrt 2 \frac{\Gamma\left( \frac{k+1}{2} \right)}{\Gamma(k/2)}\\ \sigma_k & = k-\mu_k^2 \end{align}
I must be missing something elementary, but (assuming the cited limit is correct) I can't quite see how to apply the CLT.
I've tried the delta method but didn't find it helpful here.
Added: Concerning the delta method, it seems to assume $$\sqrt k \frac{Y_k-\mu}{\sigma} \xrightarrow{d} \mathcal N(0,1)$$
and (under certain conditions on $g$) to imply $$\sqrt k \frac{g(Y_k)-g(\mu)}{ |g'(\mu)| \sigma} \xrightarrow{d} \mathcal N(0,1)$$
So if we set $Y_k = \frac{X^2_1+\cdots + X^2_k}{k}$ and $g = \sqrt{\cdot}$ then $\mu = \mathbb E(Y_k) = 1$ and $\sigma = \sqrt{ \mathbb V (Y_k) } = \sqrt 2$, so $|g'(\mu)| = 1/2$. So we end up with $$ \frac{ \chi_k - 1/\sqrt k }{ \sqrt 2 / 2 } \xrightarrow{d} \mathcal N(0,1) $$ which is quite different from the above cited from Wikipedia.