What is the limiting distribution of $\sqrt{n}(\bar{X}_n-p)-\sqrt{n}(\bar{Y}_n-q)$? For two sequence of random variables $X_1,\dots, X_n\sim_{iid} Beroulli(p) X_i$ and $Y_1,\dots, Y_n\sim_{iid} Beroulli(q)$ ($X_i$, $Y_j$ are independent), we have CLT
$$
\sqrt{n}(\bar{X}_n-p)\to N(0,p(1-p))
$$
and
$$
\sqrt{n}(\bar{Y}_n-q)\to N(0,q(1-q))
$$
Using Delta method, we get
$$
\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p}))\to N(0, \frac{1}{p(1-p)})
$$
(1) Question: What is the distribution of $\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})-\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log \frac{q}{1-q}))$?
Do we have $$\sqrt{n}(\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})-\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log \frac{q}{1-q}))\to N(0, \frac{1}{p(1-p)}-\frac{1}{q(1-q)})?$$ Otherwise, how to get it?
I think I am right.
(2) Question: what is the consistent estimator of $\frac{1}{p(1-p)}-\frac{1}{q(1-q)}$?

I'm answering the title question; the question in the body is unclear to me.

Using the rules for a variance operator we have:
$$\begin{align}
\mathbb{V}(\sqrt{n} (\bar{X}_n-p) - \sqrt{n} (\bar{Y}_n-q))
&= \mathbb{V}(\sqrt{n} (\bar{X}_n-p)) + \mathbb{V}(\sqrt{n} (\bar{Y}_n-q)) \\[14pt]
&= n \cdot \mathbb{V}(\bar{X}_n-p) + n \cdot \mathbb{V}(\bar{Y}_n-q) \\[14pt]
&= n \cdot \mathbb{V}(\bar{X}_n) + n \cdot \mathbb{V}(\bar{Y}_n) \\[8pt]
&= n \cdot \frac{p(1-p)}{n} + n \cdot \frac{q(1-q)}{n} \\[6pt]
&= p(1-p) + q(1-q). \\[6pt]
\end{align}$$
Since linear functions of normal random variables are normal random variables, you have the limiting distribution:
$$\sqrt{n} (\bar{X}_n-p) - \sqrt{n} (\bar{Y}_n-q)
\overset{\text{approx}}{\sim} \text{N}(0, p(1-p) + q(1-q)).$$
