# Convergence in distribution of $X_n^2$, where $X_n$ converges to $X$ with a uniform distribution

$X_n$ converges in distribution to $X$ with a uniform distribution on $[0,2]$. Does it imply that $X_n^2$ converges in distribution?

My attempt:

I want to show the convergence of cdfs, so:

$\Pr\left(X_n^2 \le t\right)=\Pr(-\sqrt{t}\le X_n \le \sqrt{t})\rightarrow \int_{-\sqrt{t}}^{\sqrt{t}} \frac{1}{2}\chi_{[0,2]}(x)dx=\frac{1}{2}\int_{0}^{\sqrt{t}}\chi_{[0,2]}(x)dx= 0$ for $t \le0$, $\frac{\sqrt{t}}{2}$ for $0 <t \le4$ and $1$ for $t\ge 4$.

Could you help me to determine what distribution stands for the limit cdf?

It seems to be $\mathrm{Beta}\left(\frac{1}{2},1\right)$, but I'm not sure...

Yes, $X_n^2 \overset{d}{\rightarrow}X^2$ by the Continuous Mapping Theorem.