Prove formally that $\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$ converges to $N(0,1)$ I'm not able to prove that
$$\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}$$
converges in distribution to $N(0,1)$. Here, $\hat{p}_1$ and $\hat{p}_2$ are independent estimations of the same proportion, $p$, with sample sizes $n_1$ and $n_2$, respectively. Besides,
$$\hat{p}=\frac{n_1\hat{p}_1+n_2\hat{p}_2}{n_1+n_2}.$$
I know that
$$\frac{\hat{p}_i-p}{\sqrt{\dfrac{p(1-p)}{n_i}}}$$
converges in distribution to $N(0,1)$, for $i=1,2$. My idea was to somehow subtract these two fractions to get the desired property, but convergence in distribution is not guaranteed to be preserved under substraction. I also have that
$$\text{E}(\hat{p}_1-\hat{p}_2)=0,\ \text{Var}(\hat{p}_1-\hat{p}_2)=p(1-p)\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right).$$
However, I am not able to come up with a way to apply the Central Limit Theorem or something similar.
Any help would be appreciated.
PD: sorry if this was asked before, I searched in aproach.xyz and couldn't find anything.
 A: This is not an answer but it is too long for a comment.
I suppose that $n_2$ is greater than $n_1$ (the other case should be analogous). Then, we can calculate the sample estimation $\hat{p}_2$ with $n_1$ elements instead of $n_2$. Call this (worse) estimation $\hat{p}_2'$. It is known that
$$\frac{\hat{p}_1-p}{\sqrt{\dfrac{p(1-p)}{n_1}}}$$
converges to $N(0,1)$ in probability by the Central Limit Theorem. Similarly,
$$\frac{\hat{p}_2'-p}{\sqrt{\dfrac{p(1-p)}{n_1}}}\xrightarrow{p} N(0,1).$$
Therefore, since substraction and convergence in probability behave nicely it follows that
$$\frac{\hat{p}_1-\hat{p}_2'}{\sqrt{\dfrac{p(1-p)}{n_1}}}\xrightarrow{p}N(0,1)-N(0,1)=N(0,2).$$
Getting rid of the $2$ in $N(0,2)$ we can write this as
$$\frac{\hat{p}_1-\hat{p}_2'}{\sqrt{p(1-p)\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_1}\right)}}\xrightarrow{p}N(0,1).$$
Also, since $\hat{p}_1\xrightarrow{p}p$ and $\hat{p}_2\xrightarrow{p} p$ it follows that $\hat{p}\xrightarrow{p}p$. Thus, we can multiply the fraction above by $\sqrt{p(1-p)}$ and divide it by $\sqrt{\hat{p}(1-\hat{p})}$ to obtain that
$$\frac{\hat{p}_1-\hat{p}_2'}{\sqrt{\hat{p}(1-\hat{p})\!\left(\dfrac{1}{n_1}+\dfrac{1}{n_1}\right)}}\xrightarrow{d}N(0,1),$$
by Slutsky's Theorem.
Now, the problem is that I don't know how to exchange $\hat{p}_2'$ for $\hat{p}_2$, and $n_2$ for $n_1$. This seems reasonable, but I don't know how to prove formally that the limit $n_1\to \infty,\ n_2\to \infty$ would be the same as the limit $n_1=n_2\to \infty$, which is what we have above.
