Let $(X,Y)$ be bivariate normal distributed and $$ S_X^2 = \sum_i (X_i - \bar{X})^2 \\ S_Y^2 = \sum_i (Y_i - \bar{Y})^2. $$ Now I want to derive the limit of $$ T_n:= \left ( \sqrt{n} \left ( \frac{\bar{X}}{S_X} - \frac{\mu_X}{\sigma_X} \right), \sqrt{n} \left ( \frac{\bar{Y}}{S_Y} - \frac{\mu_X}{\sigma_Y} \right) \right). $$ Then $$ \lim_{n \to \infty} T_n = N(0,\Sigma). $$ The multivariate delta method would tell me that $$ \Sigma = \begin{pmatrix} 1 +\frac{\mu_X ^2}{2 \sigma_X^2} & \rho +\frac{\rho^2 \mu_X \mu_Y}{2 \sigma_Y \sigma_X} \\ \rho +\frac{\rho^2 \mu_X \mu_Y}{2 \sigma_Y \sigma_X} & 1 +\frac{\mu_Y ^2}{2 \sigma_Y^2} \end{pmatrix} $$ while I would expect from Slutsky's theorem $$ \Sigma = \begin{pmatrix} 1 & \rho\\ \rho & 1 \end{pmatrix}. $$
Where is the error in my thinking?
So, I think I figured it out. While the application of the Slutsky theorem can give us the limits of $$ \lim_{n \to \infty} \sqrt{n} \left ( \frac{\bar{X}-\mu_X}{S_X} \right ) \\ = \lim_{n \to \infty} \sqrt{n} \left ( \frac{\bar{X}-\mu_X}{\sigma_X} \right ), $$ it cannot be used to obtain the limit of $$ \sqrt{n} \left ( \frac{\bar{X}}{S_X} - \frac{\mu_X}{\sigma_X} \right). $$ Therefore we need the Delta method and also obtain a different covariance matrix.