When $n$ approaches $\infty$, what is the probability limit of ${\sqrt n}X_n^2$ given that the probability limit of $X_n$ is $0$? When $n$ approaches $\infty$, what is the probability limit of ${\sqrt n}X_n^2$ given that the probability limit of $X_n$ is $0$? I know that by continuous mapping theorem, if $\text{plim}_{n\to\infty}X_n=0$, then, $\text{plim}_{n\to\infty}X_n^2=0$. After that, $\text{plim}_{n\to\infty}{\sqrt n}X_n^2$ would be $\sqrt\infty*0$, which I think is indeterminate, however, can I conclude this as if it was a normal limit?, is there a way to get a determinate probability limit in this case? Thank you

No. There are examples of rv.s $X_n$ for which $\mathrm{plim}_{n\to\infty}X_n=0$ and for which $\sqrt n X_n^2 \to\infty$ (in distribution, say) and other for which $\sqrt n X_n^2\to 0$, and all sorts of wild stuff in between. Just let $X_n=\pm 1/\log n$ or $X_n = \pm1 /n!$ and so on, where (if you want) the signs are chosen randomly.