I found the following exercise on an old exercise sheet of my university and I have some problems with understanding the result: Let $X_1\dots \ X_n$ be iid random variables with $X_1\sim$ $Ber(p)$ and $Y_1\dots Y_n$ be iid random variables with $Y_1\sim Poisson(p)$. Denote $ S_X=\sum_{k=1}^{n}X_k$ and $S_Y=\sum_{k=1}^{n}Y_k$. We have to check if the following holds: $$\lim_{n\to\infty} \left(P\left(\frac{S_X-E\left[S_X\right]}{\sqrt{Var\left(S_X\right)}}\leq x\right)-P\left(\frac{S_Y-E\left[S_Y\right]}{\sqrt{Var\left(S_Y\right)}}\leq x\right)\right)=0$$ I thought it to be true since the $S_X$ and $S_Y$ terms both converge to the standard normal distribution in probability i.e. $\Phi (x)$. But in class it was said to be false since the rate of convergence are the same as $1/\sqrt{n}$ and the variances differ i.e. $Var(S_X)=np(1-p)$ and $Var(S_Y)=np$. Does that really matter if we take the limit?
The teacher made the following argument: By the central limit theorem it holds that $$ \sqrt{n}\left( \frac{S_X}{n} -E[X_1]\right)\rightarrow N(0,Var(X_1))$$
$$ \sqrt{n}\left( \frac{S_Y}{n} -E[Y_1]\right)\rightarrow N(0,Var(Y_1))$$ Hence, $$\lim_{n\to\infty} \left(P\left(\frac{S_X-E\left[S_X\right]}{\sqrt{Var\left(S_X\right)}}\leq x\right)-P\left(\frac{S_Y-E\left[S_Y\right]}{\sqrt{Var\left(S_Y\right)}}\leq x\right)\right)$$ $$=\lim_{n\to\infty} \left(P\left(\sqrt{n}\left( \frac{S_X}{n} -E[X_1]\right)\leq x\cdot \sqrt{Var(X_1)}\right)-P\left(\sqrt{n}\left( \frac{S_Y}{n} -E[Y_1]\right)\leq x\cdot \sqrt{Var(Y_1)}\right)\right)\neq 0$$ But I think that the argument made by the teacher does not hold since in this case we do not compare $\Phi_X(x)$ with $\Phi_Y(x)$ but $\Phi_X(x\cdot Var(X_1))$ with $\Phi_Y(x\cdot Var(Y_1))$ and by normalizing both $\Phi_X$ and $\Phi_Y$ the difference converges to $0$. (Here $\Phi_X$ and $\Phi_Y$ are the cdfs of $N(0, Var(X_1)$ and $N(0, Var(Y_1)$ resp.)