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.
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 $$\left[\begin{matrix} 9 & -3 \\ -3 & 1 \\ \end{matrix}\right].$$