Let $A=(a_{ij}(x))$ and $B=(b_{ij}(x))$ be invertible symmetric positive definite ${n\times n}$ matrices, $a_{ij}(x)\geq 0,\,b_{ij}(x)\geq0\,\forall i,j$, also let $y$ be a $n$-vector.
Note that $Q=y^T(A- B)^{-1}y$ is a scalar.
As $x\to \infty$, $a_{ij}(x)\to 0$, $b_{ij}(x)\to 0$. Can we prove that $|Q|\to \infty$? My suspect is due to $(A-B)^{-1}=\frac{1}{\det(A-B)}\mbox{Adj}(A-B)$, so the determinant will tend to $zero$.
It turns out that even if $A - B \to 0$, we might not have $|Q| \to \infty$. For example, consider $$ A(x) = \frac 1x \pmatrix{2&0\\0&1}, \quad B(x) = \frac 1x\pmatrix{1&0\\0&2}, \quad y = \pmatrix{1\\1}. $$ We have $Q = 0$ for all $x$, so that $\lim_{x \to \infty} |Q| = 0$.
Suppose that $M(x) = A(x)-B(x)$ is positive definite (for all $x$), $M(x) \to 0$ as $x \to \infty$, and $y$ is non-zero. By the Rayleigh-Ritz theorem, we have $$ Q(x) = y^TM^{-1}(x)y \geq \lambda_{\min}(M^{-1}(x)) \|y\|^2 = [\lambda_{\max}(M(x))]^{-1}\|y\|^2 \geq \|M(x)\|^{-1}\|y\|^2. $$ Here, $\|M(x)\|$ could refer to any submultiplicative matrix norm (for instance, we can take the Frobenius norm $\|M\| = \sqrt{\sum_{i,j}m_{ij}^2}$). Since $M(x) \to 0$, we have $\|M(x)\| \to 0$. Since $\|y\| \neq 0$, it follows that $\|y\|^2/\|M(x)\| \to \infty$, so that we indeed have $Q(x) \to \infty$.