*(Note: I have asked this question before on math.stackexchangecom, but it wasn't answered, so I am trying again here).* 


The question is pretty much in the title; $X$ is an $n\times r$ matrix with $n>r$, and of course I am asking for the limit in $\epsilon \rightarrow 0$. 


The interesting case is the case in which there are some zeroes on the diagonal of $\Delta$; if there aren’t any, it is straightforward.


I am interested both by a nice expression of the limit and by a way to compute it numerically without rounding issues. 


I am aware that the limit is proportional to the comatrix of $X'\Lambda X$ where $\Lambda$ is obtained by replacing the zeroes on the diagonal of $\Delta$ by ones, and other entries by $0$, but I am hoping for something much better. Considering the $QR$ decomposition of $X$ seems a good starting point but I wasn’t able to go anywhere from this. 


**Example** 


With $X = \left[\begin{matrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{matrix}\right]$ and $\Delta = \left[\begin{matrix} 8 & 0& 0\\ 0& 2 &0 \\ 0& 0& \varepsilon \end{matrix}\right]$ it is easy to see that 
$$ (X' \Delta^{-1} X) =
 \left[\begin{matrix} 5/8 + 1/\varepsilon & 9/8 + 3/\varepsilon \\ 
 9/8 + 3/\varepsilon & 17/8 + 9/\varepsilon\end{matrix}\right] $$
and its determinant is $1/\varepsilon + 1/16$, hence the limit in $\varepsilon =0$ is
$$\left[\begin{matrix} 9 & -3 \\ -3 & 1 \\ \end{matrix}\right].$$



Limit of $(X' (\Delta + \varepsilon I_n)^{-1} X)^{-1}$ with $\Delta$ an $n \times n$ diagonal matrix
By the properties of Schur complements, your limit should be the $(1,1)$ block of
$$
\begin{bmatrix}
0 & X'\\
X & -\Delta
\end{bmatrix}^{-1}.
$$
(when it is invertible). Not sure if this counts as "much better". It also depends on what you are trying to do with it -- numerical computation? You need a closed formula in a proof?


Added remark: matrices of this kind are sometimes called "saddle-point matrices" and appear often in quadratic optimization.


