Let $A\in\mathbb{R}^{n\times n}$ be a matrix with eigenvalues having (strictly) negative real part. Let $X\in\mathbb{R}^{n\times n}$, $X\succ 0$, be a positive definite matrix and let $P\succ 0$ be the (unique) solution of the following Lyapunov matrix equation
$$\tag{1}\label{eq1}
AP+PA^\top = - X.
$$


> **My question.** Let $\{X_n\}_{n\ge 0}$, $X_n\succ 0$, be a sequence of positive definite matrices, and let $\{P_n\}_{n\ge 0}$, $P_n\succ 0$, be the corresponding sequence of solutions of \eqref{eq1}. Suppose that $\lim_{n\to \infty} X_n=\bar{X}\succeq 0$ and  $\lim_{n\to \infty} P_n=\bar{P}\succeq 0$ is *singular*. I'm wondering whether $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ always converges to a finite matrix.

A few remarks are in order. 

 1. If $\bar{P}$ is singular, then, $\bar{X}$ must be singular as well.
 2. If $A$ is a scalar matrix, i.e. $A=\alpha I$, $\alpha<0$, then it is quite easy to see that $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ converges to a finite matrix.
 3. By pre- and post-multiplying \eqref{eq1} by $P^{-1/2}$, it follows that if $\lim_{n\to \infty} P_n^{-1/2}A P_n^{1/2}$ is finite, then $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ is finite as well.