Delta-method for the convergence in distribution of $a_n\left(h(\bar X_n) - h(\theta) - b_n\right)$ when $h'(\theta)=0$ 
Suppose that $X_i$ are i.i.d. following a Bernoulli distribution with $\mathbb{P}(X_i = 1) = \theta$. Let $\bar X_n$ be the sample mean of $X_1, ..., X_n$. Let $h(t) = t^3 - t$. Find $a_n$ and $b_n$ such that $a_n\left(h(\bar X_n) - h(\theta) - b_n\right)$ converges in distribution (to a non-degenerate limit).

Notice that $h'(t) = 3t^2 - 1$ and has zeros at $\pm \frac{1}{\sqrt{3}}$, so if $\theta = \frac{1}{\sqrt{3}}$ (which is plausible), $h'(\theta) = 0$.
Assume that is not the case. Then $\bar X_n$ satisfies the delta theorem and a simple application of the theorem yields:
$$\sqrt{n}\left[h(\bar X_n) - h(\theta)\right] \to N\left(0, \theta(1 - \theta)\left(h'(\theta)\right)^2\right)$$
Thus pick $a_n = \sqrt{n}$ and $b_n = 0$ and the problem is solved.
But if $\theta = \frac{1}{\sqrt{3}}$, the delta theorem cannot be applied.
Thus my question is: do I need to find a new $a_n$ and $b_n$ for the case $\theta = \frac{1}{\sqrt{3}}$?
 A: The so-called delta-method stems from two basic results: 


*

*By the usual CLT, $\sqrt{n}(\bar X_n-\theta)\to\sigma_\theta Z$ in distribution, where $Z$ is standard normal and $\sigma_\theta^2=\mathrm{var}(X_1)$.

*If $h'(\theta)\ne0$, then $h(x)-h(\theta)\sim h'(\theta)(x-\theta)$ when $x\to\theta$.


Putting 1. and 2. together, one sees that, if $h'(\theta)\ne0$, then $h(\bar X_n)-h(\theta)\sim h'(\theta)(\bar X_n-\theta)$, which implies:

$\sqrt{n}(h(\bar X_n)-h(\theta))\to a_\theta Z$ in distribution with $a_\theta=h'(\theta)\sigma_\theta$.

This is the result you recall since $a_\theta Z$ is centered normal with variance $a_\theta^2=h'(\theta)^2\theta(1-\theta)$.
Likewise, if $h'(\theta)=0$, our point 2. above becomes:


*If $h'(\theta)=0$ and $h''(\theta)\ne0$, then $h(x)-h(\theta)\sim \frac12h''(\theta)(x-\theta)^2$ when $x\to\theta$.


Thus, if $h'(\theta)=0$ and $h''(\theta)\ne0$, then $h(\bar X_n)-h(\theta)\sim \frac12h''(\theta)(\bar X_n-\theta)^2$, which implies: 

If $h'(\theta)=0$, then $n(h(\bar X_n)-h(\theta))\to b_\theta Z^2$ in distribution with $b_\theta=\frac12 h''(\theta)\sigma_\theta^2$.

It remains to identify the distribution of $b_\theta Z^2$ if $Z$ is standard normal. Can you do that?
