Show that $\sqrt{n}(\ln(2\overline{X}_{n}-\ln(\theta))\overset{\mathcal{D}} \to N(0,3^{-1})$ 
Let $(X_n)$ be a sequence of independent, identically distributed random variables, with $X_n \sim \mathcal{U}[0,\theta]$ for some $\theta > 0$.  Show that $\sqrt{n}(\ln(2\overline{X}_{n})-\ln(\theta))\overset{\mathcal{D}} \to N(0,3^{-1})$.

Im not being able to solve this problem, Im stucked in the central limit theorem with
$$\sqrt{n}\frac{\overline{X}_{n}-\frac{\theta}{2}}{\frac{\theta}{\sqrt{12}}}\overset{\mathcal{D}} \to N(0,1)$$
And I've already proved various results regarding continuous functions and convergence in distribution. But Im not being able to take the last formula to the one in the problem. Can you just give me a hint?
 A: Have a look at the Delta Method.
A: $\DeclareMathOperator{\Pr}{\mathbb{P}}$
Let $Z_n = \dfrac{X_n}{\theta}$ then $Z_n \sim \mathcal{U}[0,1]$, and let $S_n = Z_1 + \dots + Z_n.$
Choose any real $t$, we have
$
\begin{split}
\Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)) \le t) = \Pr(\sqrt{n}(\ln(2\frac{\bar{X}_n}{\theta}) ) \le t)\\ =\Pr(\sqrt{n}(\ln(2nS_n) ) \le t) \\=\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1\right)\right)\\
=\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3} +\sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right)\right)
\end{split}
$
Since $\exp(x) \geq 1 + x$ for all real $x$ we have   $0 \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right) = \sqrt{3n}(\dfrac{t^2}{2n} + \text{O}(\dfrac{1}{n\sqrt{n}})) = \text{O}(\dfrac{1}{\sqrt{n}})$.
So for sufficiently large $n$ and any $ \epsilon > 0$ we have $0 \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right) \leq \epsilon$
so $\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3}\right) \leq \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)) \le t) \leq \Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3} + \epsilon\right).$
So letting $n \to \infty$ and using CLT we have
$$\Phi(t\sqrt{3}) \leq \lim_{n\to\infty} \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)\leq t)  \leq \Phi(t+\epsilon),$$ where $\Phi$ is the CDF of the standard normal distribution. Since $\epsilon > 0$ artibtrary and $\Phi$ is continuous we have $\lim_{n\to\infty} \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta) \leq t) = \Phi(t\sqrt{3})$ and we are done.
