Let $X_1, ... X_n \sim $ Unif$(0,\theta)$ and $X_{(n)} = \max X_i$
Find the limiting distribution of $\displaystyle \frac{(2 \sum_{i=1}^n X_i - \theta)}{\sqrt{n}X_{(n)}}$
I know I should be using Slutskys because $X_{(n)} \to \theta$ in probability.
It seems that maybe I should re-write this as $\displaystyle \sqrt{n}\frac{(2 \sum_{i=1}^n X_i - \theta)}{nX_{(n)}} = \sqrt{n}\frac{(2 \frac{1}{n} \sum_{i=1}^n X_i - \theta/n)}{X_{(n)}} $
Then we have that $\frac{1}{n} \sum_{i=1}^n X_i \to \mu = \frac{\theta}{2}$ by the Law of Large Numbers.
But then in the limit what we end up with is $\sqrt{n} \frac{(2 \frac{\theta}{2}-\theta)}{\theta} = 0$
So it seems I am doing this incorrectly.
EDIT: Using the comment below I can re-write as
$\sqrt{n}\frac{2 (\frac{1}{n} \sum_{i=1}^n X_i - \theta/2n)}{X_{(n)}}$