How to calculate $ \lim_{\alpha \to 1} (I-\alpha A)^{-1}( I- A)$? Suppose that for all $\alpha \in (0,1)$, the matrix $\mathbf I-\alpha \mathbf A$ is invertible, but that $\det(\mathbf I-\mathbf A)=0$. How do I calculate
$$ \lim_{\alpha \to 1} (\mathbf I-\alpha \mathbf A)^{-1}(\mathbf I- \mathbf A)?$$

In the special case where $A_{n \times n}$ is a real symmetric matrix the limit can be explicitly computed in terms of the spectral decomposition of $A$.
The condition $\det(I-A)=0$ ensures $1$ is an eigenvalue of $A$. 
The condition $\det(I-\alpha A) \neq 0$ for $\alpha \in (0,1)$ implies $\det(\frac{1}{\alpha}I -A) \neq 0$ for $\alpha \in (0,1)$, i.e., $\det(\mu I-A) \neq 0$ for $\mu > 1$. So all eigenvalues of $A$ that are distinct from $1$ (if any) must lie in the interval $(-\infty,1)$. 
Let $r > 0$ be the multiplicity of $1$ as an eigenvalue of $A$, and let $\lambda_{r+1},\dots,\lambda_{n}$ denote the remaining eigenvalues, with each eigenvalue repeated as many times as its multiplicity.
We can find an orthronormal basis of $\mathbb{R}^n : \{u_1,u_2,\dots,u_r,u_{r+1},\dots,u_n\}$ such that 
$$\label{e:1}A = \sum_{i=1}^{r}u_i u_i^T + \sum_{i=r+1}^{n}\lambda_iu_iu_i^T \tag{*}$$ 
and the orthonormality implies 
$$
\label{e:2}
I = \sum_{i=1}^{n}u_iu_i^T \tag{+}.
$$
We have from $\eqref{e:1}$ and $\eqref{e:2}$ $$I-A = \sum_{i=r+1}^{n}(1-\lambda_i)u_iu_i^T$$ and
$$
I - \alpha A = \sum_{i=1}^r (1 - \alpha)u_iu_i^T + \sum_{i=r+1}^{n}(1-\alpha\lambda_i)u_iu_i^T.
$$
Note for $\alpha \in (0,1)$ and $i > r$ we have $1 -\alpha \lambda_i \neq 0$ so
$$
(I-\alpha A)^{-1} = \frac{1}{1-\alpha}\sum_{i=1}^ru_i u_i^T + \sum_{i=r+1}^n\frac{1}{1-\alpha\lambda_i} u_iu_i^T
$$
and so
$$
(I-\alpha A)^{-1}(I - A) = \sum_{i=r+1}^{n}\frac{1-\lambda_i}{1-\alpha\lambda_i}u_iu_i^T.
$$
So,
$$
\lim_{\alpha \to 1^{-}}(I-\alpha A)^{-1}(I - A) = \sum_{i=r+1}^{n}u_iu_i^T.
$$
