Asymptotic normality of MLE in exponential with higher-power x Given the distribution:
$f(x;\theta) = \frac{3}{\theta}x^2e^{-x^3/\theta}$ if $x>0$
the MLE for $\theta$ is $\frac{1}{n}\sum_{i=1}^n x_i^3$. It's an unbiased estimator with variance $\theta^2/n$. The Fisher information number is $1/\theta^2$.
Now, I am supposed to check whether or not this estimator shows asymptotic normality. The problem here is that my textbook has a condition which I can't find in any other source:
"$\frac{d^2}{d\theta^2} \ln f(x;\theta)$ is a continuous function in $\theta$, uniformly in $x$."
It's this uniformly continuity that I can't seem to find anywhere else. This seems to be the only condition not met for asymptotic normality. I've found that $\frac{d^2}{d\theta^2} ln f(x;\theta) = \frac{1}{\theta^2} - 2\frac{x^3}{\theta^3}$. Since uniform continuity is equivalent with the boundedness of the derivative, I derived this to $x$ to find $ - 6\frac{x^2}{\theta^3}$, which is not bounded for x. 
My textbook also mentions that "these conditions are in particular fulfilled for (identifiable) members of the exponential class". So, my question is, is this really a necessary condition, and am I right that this MLE is not asymptotically efficient?
Thanks.
 A: So that the question does not remain open:
Set $X^3 = Z \Rightarrow X = Z^{1/3}$. Then
$$f_Z(z) = \left|\frac {\partial X}{\partial Z}\right|\cdot f_X(z^{1/3}) = \frac 13z^{-2/3}\cdot \frac{3}{\theta}z^{2/3}e^{-z/\theta} = \frac 1{\theta}e^{-z/\theta}$$
So we see that $Z=X^3$ is a random variable that follows the exponential distribution with $E(Z) = \theta$ and $\operatorname {Var}(Z) = \theta^2$.
Then 
$\hat \theta_{MLE} = \frac{1}{n}\sum_{i=1}^n z_i$ is the sample mean of $n$ (i.i.d. as clarified in the comments) exponential random variables, with $E[\hat \theta_{MLE}]=\theta$ and $\operatorname {Var}(\hat \theta_{MLE}) = \theta^2/n$ indeed.
The variance of each r.v. involved is bounded and so the Lindeberg-Levy CLT applies, meaning
$$\sqrt n(\hat \theta_{MLE}-\theta) \rightarrow_d N(0,\theta^2)$$
and the MLE is asymptotically normal.
The Fisher Information of $\theta$ is, indeed,
$$\mathcal{I}(\theta)=\operatorname{E} \left[\left. \left(\frac{\partial}{\partial\theta} \log f_Z(z;\theta)\right)^2\right|\theta \right] = \operatorname{E} \left[\left. (-\frac 1{\theta} + \frac {z}{\theta^2} \right)^2\right] = \frac 1{\theta^2}$$
So 
$$\operatorname {Avar}(\hat \theta_{MLE}) = \frac 1{\mathcal{I}(\theta)}$$
and the MLE is asymptotically efficient.
