Question on limiting variance in example 10.1.8 (Casella, Statistical Inference, 2nd edition)

Question

This is from George Casella textbook statistical inference (page 470) Example 10.1.8. Limiting variances. It says for the mean $\bar{X_n}$ of n iid normal observations with $EX=\mu$ and $VarX=\sigma^2$, if we take $T_n=\bar{X_n}$, then $lim\sqrt{n}Var\bar{X_n}=\sigma^2$ is the limiting variance of $T_n$.

I feel very confused. I think $\bar{X_n}$ is $n(\mu, \sigma^2/n)$. Then $lim\sqrt{n}*Var\bar{X_n}=lim\sqrt{n}*\sigma^2/n=\sigma^2lim\frac{\sqrt{n}}{n}=0$. The limit is 0, not $\sigma^2$.

Please point my mistake.

Answer 1

This is a typo in the book. I have the same edition as you and upon opening my copy I see that I have already made this correction. We already know that $\mathsf{Var}\bar X_n=\sigma^2/n$. Using Casella's definition of limiting variance in definition 10.1.7 we can infer the sentence in Example 10.1.8 should read:

...if we take $T_n=\bar X_n$, then $\lim n\mathsf{Var}\bar X_n=\sigma^2$ is the limiting variance of $T_n$.

Indeed, using the same notation in the definition we have $k_n=n$, $T_n=\bar X_n$, and $\tau^2=\sigma^2$.