Obtaining a Non-Singular Matrix from a Singular one by Perturbation In a paper "http://www.math.cornell.edu/~nussbaum/papers/08-1.pdf" (page 264 Lemma 2) I encountered the following way of obtaining an invertible (non-singular) matrix from a non-invertible (singular) positive semi-definite one. Let matrix $A\geq0$ be non-invertible or singular and $I$ be the identity of the same order. Then we have:
$$\lim_{x\to 0}A(A+xI)^{-1}=I$$
Why can such a limit be defined for a non-invertible matrix? 
 A: What if $A$ were the zero matrix? Then your limit would be the zero matrix. So your claim doesn't seem true, unless perhaps we make more assumptions on $A$.
A: The assertion is never true.  If $A$ is a singular $n \times n$ matrix, there is a nonzero vector $v$ such that  $A v = 0$.  Then $A (A + xI)^{-1} v = (A + xI)^{-1} A v = 0$, so this can't converge to $Iv = v$ as $x \to 0$.
What is true is that if $0$ is not defective as an eigevalue of $A$, i.e. the algebraic and geometric multiplicities of $0$ as an eigenvalue of $A$ are equal, then the limit exists and is a projection on the range of $A$.
EDIT: The statement in the paper you referenced is patently wrong as written, but I believe the lemma can be rescued.  In this case $A$ is a hermitian matrix, so there are no defective eigenvalues, and the integral representation of $A^t$ for $0 < t \le 1$ is still correct.  It implies
$A^t v = 0$ for $v$ in the null space of $A$, which is indeed what you want.  The case $t=0$ might be somewhat problematic, and should be dealt with separately (what do you mean by $0^0$ anyway?).
EDIT: For the proof, if $0$ is not defective, then we can write the
vector space $A$ is acting on as $V = K \oplus R$, $K = \ker A$
 the kernel or null space, $R = \text{ran}\; A$ the range.  $K$ and $R$ are both invariant under $A$, so it suffices to look separately at what happens on $K$ and on $R$.  On $K$ of course we have $A (A+xI)^{-1} = 0$, as mentioned above.   Since the restriction $A_R$ of $A$ to $R$ is invertible, $(A_R + xI)^{-1} \to A_R^{-1}$ as $x \to 0$, and thus 
$A (A + xI)^{-1} v \to v$ for $v \in R$.
A: What is true is that $\det(A+xI)$ is a polynomial having a finite number of zeros. So for $x \neq 0$ small enough, $A+xI$ is invertible.
However, that doesn't mean that $A(A+xI)^{-1}=I$.
