Find asymptotic distribution of $l(\hat{\theta_n}) - l(\theta_0)$ Where ${\hat{\theta_n}}$ are the asymptotic efficient roots for $\frac{\partial}{\partial \theta} l(\theta,\underline{X}) = 0$
I have really no clue how to begin this, but I was given
Hint: a good first step/main issue is to figure out doing a Taylor Expansion on which quantity.
I also realize that we're determining the log-likelihood ratio, and there's pretty good practical value to statistical hypothesis testing, but beyond taking the taylor expansion and doing the central limit theorem, I'm confused. Any help or guidance is really appreciated, thanks.
Update: okay so I thought about doing the taylor expansion of $l(\hat{\theta_n})$ about $l(\theta_0)$ and I get $l(\hat{\theta_n}) - l(\theta_0) = l'(\theta_0)(\hat{\theta_n} - \theta_0) + \frac{l''(\theta_0)}{2}(\hat{\theta_n} - \theta_0)^2 + \frac{l'''(\theta_m)}{6}(\hat{\theta_n} - \theta_0)^3$ where $\theta_m$ is between $\theta_n$ and $\theta_0$ but I don't seem to know where to go from here since if $\hat{\theta_n}$ converges in probability to $\theta_0$, the entire right  hand side is 0.
I thought about doing the Taylor Expansion of $l(\theta_0)$ about $\hat{\theta_n}$ since we know $\hat{\theta_n}$ is a root of $\frac{\partial}{\partial \theta} l(\theta,\underline{X}) = 0$ but I run into a similar problem as above.
 A: I actually think I solved it... Had to use the Taylor Expansion of $l(\hat{\theta_0})$ about $\hat{\theta_n}$ which is $l(\theta_0) = l(\hat{\theta_n}) + l'(\hat{\theta_n})(\theta_0 - \hat{\theta_n}) + \frac{l''(\hat{\theta_n})}{2}(\theta_0 - \hat{\theta_n})^2 + \frac{l'''(\tilde{\theta})}{6}(\theta_0-\hat{\theta_n})^3. $ 
Then, you use an assumption to show that $\frac{l'''(\tilde{\theta})}{6}(\theta_0-\hat{\theta_n})^3$ converges to $0$ in distribution, and since $\hat{\theta_n})$ is a root of $l'(\theta) = 0$, then that term goes to $0$, and you are left with $-2(l(\hat{\theta_n})-l(\theta_0)) = -l''(\hat{\theta_n})(\hat{\theta_n}-\theta_0)^2 $.
Then you do some neat algebra to show $-2(l(\hat{\theta_n})-l(\theta_0)) = -(\frac{l''(\hat{\theta_n})}{n})(\sqrt{n}(\hat{\theta_n}-\theta_0))^2 $
We know $\sqrt{n}(\hat{\theta_n}-\theta_0)$ converges to $N(0,\frac{1}{I(\theta_0)}) $ and we can show that $-\frac{1}{n}l''(\hat{\theta_n})$ converges to $I(\theta_0)$ (obviously missing a few steps here).
Thus, $l(\hat{\theta_n}) - l(\theta_0)$ is asymptotically distributed as $-\frac{\chi_1^2}{2} $
For those following along at home, this is, indeed, Wilks' Theorem.
