# Asymptotic distribution of a linear combination

If $$X_1$$ and $$X_2$$ are independent and follow asymptotic standard normal distribution as $$\min(n_1,n_2)\to\infty$$, how do I show that $$\frac{\sqrt {n_1}X_1+\sqrt {n_2}X_2}{\sqrt{n_1+n_2}}$$ also has an asymptotic standard normal distribution?

The problem would be trivial if $$n_1$$ and $$n_2$$ would be constants and $$X_1$$ and $$X_2$$ would have asymptotic normal distribution based on some other index going to $$\infty$$, because in that case $$\sqrt{n_1}X_1$$ would converge to asymptotic $$\mathcal N(0,n_1)$$ distribution, $$\sqrt{n_2}X_2$$ would converge to asymptotic $$\mathcal N(0,n_2)$$ distribution, so their sum (by independence) would converge to asymptotic $$\mathcal N(0,n_1+n_2)$$ distribution, which is what we want to show.

The same logic would also work if $$n_1/n_2$$ would be converging to a non-zero constant, because we can dividing numerator and denominator by $$n_2$$ then.

However, in this case $$n_1$$ and $$n_2$$ are not fixed, and I don't know what to do.

Write $$a_n=\sqrt{n_1}/\sqrt{n_1+n_2}$$ and $$b_n=\sqrt{n_2}/\sqrt{n_1+n_2}$$, and $$Z_n$$ for your linear combination.
Both $$a_n$$ and $$b_n$$ are bounded above by 1 and below by 0, so for every sequence $$n_i$$ there is a subsequence $$n_{i_j}$$ such that $$(a_{n_{i_j}},b_{n_{i_j}})$$ converges (to $$(a,b)$$, say). Along this subsubsequence $$Z_n\stackrel{d}{\to}N(0,1)$$.
So, for every sequence $$n_i$$ there is a subsequence $$n_{i_j}$$ such that $$Z_{n_{i_j}}\stackrel{d}{\to}N(0,1)$$. And that implies $$Z_n\stackrel{d}{\to}N(0,1)$$ (eg, see this answer)