Limiting Distribution for UMVUE I have been given this question as a homework problem. Let $X\sim Bin(p,n)$ and let $g(p)=p(1-p)$, the UMVUE is $\delta=\frac{X(n-X)}{n(n-1)}$. Determine the limit distribution of $\sqrt{n}\bigg[\delta-p(1-p)\bigg]$ when $g^{'}(p)\neq  0.$
My attempt: 
Since $X\sim Bin(p,n)$ then $X\overset{d}=\sum_{i}^{n}X_i$ where $X_i\sim Bern(p)$
Then by the CLT we know: 
$$\sqrt{n}(\bar{X}-p)\overset{d}\rightarrow N(0,p(1-p))$$
Then $\delta=\frac{n\bar{X}(1-\bar{X})}{n-1}$ by delta method it can be shown that 
$$\sqrt{n}\bigg[\delta-\frac{n}{n-1}p(1-p)\bigg]\overset{d}\rightarrow N(0,p(1-p)(\frac{n-2np}{n-1})^2)$$
I am  double guessing myself here because I believe it should go to $N(0,p(1-p)(1-2p)^2)$  since $\frac{n-2np}{n-1}\rightarrow (1-2p)$
since $\delta(p)=\frac{n}{n-1}(p(1-p))$ whose derivative does not equal zero by assumption since $\delta(p)=\frac{n}{n-1}g(p)$. 
One can conclude that since $\frac{n}{n-1}\rightarrow 1$ that 
$$\sqrt{n}\bigg[\delta-p(1-p)\bigg]\overset{d}\rightarrow N(0,p(1-p)(\frac{n-2np}{n-1})^2)$$ 
Again I believe that it should go to $N(0,p(1-p)(1-2p)^2)$
Is my logic correct? I know that the asymptotic distribution should not include n, so is my second guess correct? 
 A: Update:
Alternatively, we may apply Slutsky's theorem. https://en.wikipedia.org/wiki/Slutsky%27s_theorem
From the CLT, we have
$$\sqrt{n}(\bar{X} - p) \overset{d}\rightarrow N(0, p(1-p)).$$
From the delta method, we have
$$\sqrt{n}[\bar{X}(1-\bar{X}) - p(1-p)] \overset{d}\rightarrow N(0, p(1-p)(1-2p)^2). $$
Denote $Y_n = \sqrt{n}[\bar{X}(1-\bar{X}) - p(1-p)]$, $W_n = \frac{\sqrt{n}}{n-1}\bar{X}(1-\bar{X})$
and $Z_n = \sqrt{n}(\delta - p(1-p))$. Then $Z_n = Y_n + W_n$.
Clearly, $W_n \overset{p}\rightarrow 0$. From Slutsky's theorem, we know that
$Z_n \overset{d}\rightarrow N(0, p(1-p)(1-2p)^2)$.
Previously written
From the CLT, we have
$$\sqrt{n}(\bar{X} - p) \overset{d}\rightarrow N(0, p(1-p)).$$
From the delta method, we have
$$\sqrt{n}[\bar{X}(1-\bar{X}) - p(1-p)] \overset{d}\rightarrow N(0, p(1-p)(1-2p)^2). \tag{1}$$
Denote $Y_n = \sqrt{n}[\bar{X}(1-\bar{X}) - p(1-p)]$
and $\sigma^2 = p(1-p)(1-2p)^2$. (1) means that for each fixed real number $z$,
$$\lim_{n\to \infty} \mathrm{Pr}(Y_n \le z) = \Phi(\tfrac{z}{\sigma}) \tag{2}$$
where $\Phi(z)$ is the standard normal CDF.
Denote $Z_n = \sqrt{n}(\delta - p(1-p))$. Let us prove that for each fixed real number $z$,
$$\lim_{n\to \infty} \mathrm{Pr}(Z_n \le z) = \Phi(\tfrac{z}{\sigma}).$$
To this end, by noting that $Z_n = Y_n + \frac{\sqrt{n}}{n-1}\bar{X}(1-\bar{X})$, we have
$$\mathrm{Pr}(Z_n \le z) = \mathrm{Pr}\Big(Y_n \le z - \tfrac{\sqrt{n}}{n-1}\bar{X}(1-\bar{X})\Big).$$
Note that $0\le \bar{X}(1-\bar{X}) \le \frac{1}{4}(\bar{X} + 1 - \bar{X})^2 = \frac{1}{4}$.
We have
$$\mathrm{Pr}\Big(Y_n \le z - \tfrac{\sqrt{n}}{n-1}\tfrac{1}{4}\Big) \le \mathrm{Pr}(Z_n \le z) \le \mathrm{Pr}(Y_n \le z). \tag{3}$$
From (2), we have, for any $N > 2$, there exists $n_0(N) > N$ such that for any $n > n_0(N)$,
\begin{align}
\Phi\Big(\frac{z - \tfrac{\sqrt{N}}{N-1}\tfrac{1}{4}}{\sigma}\Big) - \frac{1}{N}
&\le \mathrm{Pr}\Big(Y_n \le z - \tfrac{\sqrt{N}}{N-1}\tfrac{1}{4}\Big)\\
&\le \mathrm{Pr}\Big(Y_n \le z - \tfrac{\sqrt{n}}{n-1}\tfrac{1}{4}\Big)\\ 
&\le  \mathrm{Pr}(Y_n \le z). \tag{4}
\end{align}
Since $\lim_{N\to \infty} \Phi\Big(\frac{z - \tfrac{\sqrt{N}}{N-1}\tfrac{1}{4}}{\sigma}\Big) - \frac{1}{N} = \Phi(\tfrac{z}{\sigma})$,
by using (2), (4) and the squeeze theorem, we have 
$$\lim_{n\to \infty} \mathrm{Pr}\Big(Y_n \le z - \tfrac{\sqrt{n}}{n-1}\tfrac{1}{4}\Big) = \Phi(\tfrac{z}{\sigma}). \tag{5}$$
From (2) and (5), by applying the squeeze theorem for (3), we have
$$\lim_{n\to \infty} \mathrm{Pr}(Z_n \le z) = \Phi(\tfrac{z}{\sigma}) \tag{6}.$$
(6) means that
$$\sqrt{n}(\delta - p(1-p)) \overset{d}\rightarrow N(0, p(1-p)(1-2p)^2).$$
We are done.
