Find a function such that follows to normal in distribution Suppose that $X_{n}\sim \text{Binomial}(n,\theta)$, where $n=1,2,\ldots$ and $0<\theta<1$. Find a function $g$ such that $\sqrt{n}(g(\frac{1}{n}X_n)-g(\theta))\xrightarrow{D} N(0,1)$ for each value of $\theta\in(0,1)$. Can someone give me hints by using Delta method? I was trying to prove $\sqrt{n}(X_{n}-\theta)\xrightarrow{D} N(0,\sigma^2)$. But I find I only know $\bar{X}_n$ has similar property.
Delta Method Theorem
Let $Y_n$ be sequence of random variables that satisfies $\sqrt{n}(Y_n-\theta)\xrightarrow{D}N(0,\sigma^2)$. For a given function $g$ and a specific value of $\theta$, suppose that $g'(\theta)$ exists and is not $0$. Then $$\sqrt{n}[g(Y_n)-g(\theta)]\xrightarrow{D}N(0,\sigma^2[g'(\theta)]^2).$$
 A: To find such $g$ we shall use the following lemma:
Lemma: if $\{T_n\}_{n=1}^{\infty}$ is a sequence of random variables such that
$$\sqrt{n}(T_n-\theta) \longrightarrow N(0,\sigma)$$
and $g$ is a derivable function from $\mathbb{R}$ to $\mathbb{R}$ so that $g'(\theta)\not=0$, then
$$\sqrt{n}(g(T_n)-g(\theta)) \longrightarrow N(0,|g'(\theta)|\sigma)$$
You may find this result proved in most of the classical statistical inference books.
In our case, we have that $Y_n=\frac{1}{n}X_n$ is a $B(1,\theta)$ (this follows from the fact that $X_n=nY_n$, and that can only happen if each $Y_n$ is $B(1,\theta))$
Knowing this, and applying the previous result, we must find $g$ so that
$$|g'(\theta)|\sqrt{\theta(1-\theta)}=1$$
Because $\sigma$ represents the typical deviation of the population, in this case, $\sigma = \sqrt{\theta(1-\theta)}$
We have two functions that satisfy the following equation, namely
$$g_1(\theta)=\int \frac{d\theta}{\sqrt{\theta-\theta^2}}\qquad \& \qquad g_2(\theta) = -\int \frac{d\theta}{\sqrt{\theta-\theta^2}}$$
Which both have the following expression
$$g_1(\theta)=arsin(2\theta-1)+C\quad\&\quad g_2(\theta)=arccos(2\theta-1)+C\quad C\in\mathbb{R}$$
The step by step calculation of $g_1(\theta)$ is presented below (the other one is analogous):
$$\begin{align} \int \frac{d\theta}{\sqrt{\theta-\theta^2}} & =\int \frac{d\theta}{\sqrt{\frac{1}{4}-\big(\theta-\frac{1}{2}\big)^2}}=\int \frac{d\theta}{\frac{1}{2} \sqrt{1-\bigg(\frac{\theta-\frac{1}{2}}{\frac{1}{2}}\bigg)^2}}=\int \frac{2d\theta}{\sqrt{1-(2\theta-1)^2}} \\\\ & = arcsin(2\theta+1)+C\quad C\in \mathbb{R}\end{align}$$
As it is an immediate integral
It is immediate to verify that both functions satisty the regularity conditions given above: namely, that $0<\theta<1$ and $|g'(\theta)|\not=0$ for all $\theta$
