Let $X_1,\dots,X_n$ be an iid sample from a uniform distribution $U[-\theta,\theta]$ for some parameter $\theta>0$. Define $X_{(n)}=\max\{X_i: 1\leq i\leq n\}$ and to estimate $\theta$ consider the estimator $T_n=\sqrt{\frac{3}{n}\sum_{i=1}^nX_i^2}$.
I have to determine the limit distribution of the sequence $\sqrt{n}(T_n-\theta)$. My approach is as follows. I know from the law of large numbers that $\overline{X_n^2}=\frac{1}{n}\sum_{i=1}^nX_i^2\to E(X_1^2)=Var(X_1)=\frac{\theta^2}{3}$.
So $\sqrt{n}(\overline{X_n^2}-\frac{\theta^2}{3})\to N(0,Var(X_1^2))$ in distribution by the central limit theorem, where $Var(X_1^2)=E(X_1^4)-(E(X_1^2))^2=\frac{\theta^4}{30}$.
So by the delta method we have, with $\phi(x)=\sqrt{3x}$, that $\sqrt{n}(\phi(\overline{X_n^2})-\phi(\frac{\theta^2}{3}))=\sqrt{n}(T_n-\theta)\to \phi'(\frac{\theta^2}{3})N(0,\frac{\theta^4}{30})=\frac{3}{2\theta}N(0,\frac{\theta^4}{30})=N(0,\frac{3\theta^2}{40})$ in distribution.
I don't know for sure if this is correct. Can somebody tell me if this is the right approach?
EDIT:
I made some mistakes in the calculation.
$Var(X_1^2)=E(X_1^4)-(E(X_1^2))^2=\frac{\theta^4}{5}-\frac{\theta^4}{9}=\frac{4\theta^4}{45}$, hence
$\sqrt{n}(\phi(\overline{X_n^2})-\phi(\frac{\theta^2}{3}))=\sqrt{n}(T_n-\theta)\to \phi'(\frac{\theta^2}{3})N(0,\frac{4\theta^4}{45})=\frac{3}{2\theta}N(0,\frac{4\theta^4}{45})=N(0,\frac{9}{4\theta^2}\frac{4\theta^4}{45})=N(0,\frac{\theta^2}{5})$ in distribution.
This answer makes some more sense, my guess is that this is correct.