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http://en.wikipedia.org/wiki/Delta_method
In the Wikipedia article, it was assumed that $g'(\theta)$ must exist and that $g'(\theta)$ is non-zero valued.
Is it possible to find the asymptotic distribution for $\sqrt{n}(g(X_n)-g(\theta))$
given that $g'(\theta)$ might be zero and $\sqrt{n}(X_n-\theta) \stackrel{d}{\rightarrow} N(0,\sigma^2)$?

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When the first derivative is 0 you can use the second-order delta method.

Loosely:

$g(X_n)=g(\theta+X_n-\theta)=g(\theta)+(X_n-\theta)g'(\theta)+(X_n-\theta)^2/2\cdot g''(\theta)+o_p(1)$

So $g(X_n)-g(\theta)= \frac{g''(\theta)}{2} (X_n-\theta)^2 +o_p(1)$

But $n(X_n-\theta)^2/\sigma^2\stackrel{d}{\to} \chi^2_1$

... and so forth.

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