limit of vector sequence is in range of limit of matrix sequence $\{A_k\}_{k=1}^{\infty}$ is a convergent sequence of real $n \times d$ matrices of rank $n$, and the limit $\lim_{k \to \infty}A_k := A_*$ also has rank $n$.
$\{x_k\}_{k=1}^{\infty}$ is a convergent sequence of real $d$ dimensional vectors, $\lim_{k \to \infty}x_k := x_*$.
We are given that $x_k \in \text{Range}(A_k^T)$ for all $k$. Is it true that $x_* \in \text{Range}(A_*^T)$?
 A: Here is another way of seeing that the answer is "yes".
Recall that the orthogonal projection onto $\mathrm{Range}(A_k^T)$ is $P_k = A_k^T(A_kA_k^T)^{-1}A_k$. Since $A_k$ has full rank $n$ for each $k$, the inverse in the middle always exists, and since the limit $A_*$ also has full rank $n$ that inverse does not blow up (i.e., the singular values of $A_kA_k^T$ are bounded away from $0$). Thus $\lim_{k\rightarrow \infty}P_k = P_* = A_*^T(A_*A_*^T)^{-1}A_*$ is the orthogonal projection onto $\mathrm{Range}(A_*^T)$, so
$$
\|P_*x_* - x_*\| = \lim_{k\rightarrow \infty} \|P_k x_k - x_k\| = \lim_{k\rightarrow\infty}0 = 0,
$$
so $P_*x_* = x_*$, so $x_* \in \mathrm{Range}(A_*^T)$.
A: Yes: this is equivalent to the following fact: if $B_k = \begin{bmatrix}x_k^T \\ A_k\end{bmatrix}$ does not have full rank for each $k$, then the limit does not have full rank either.
In turn, this follows from the interpretation of rank in terms of determinants: a matrix $B_k \in \mathbb{R}^{(n+1)\times d}$ is singular iff the determinant of each of its $(n+1)\times (n+1)$ submatrices vanishes.
