Let $A=(a_{ij})$ be a invertible symmetric positive definite matrix, $a_{ij}>0\,\,\forall i,j$, let $B=A^{-1}=(b_{ij})$. How do I prove, if it is really true, whether $a_{ij}\approx 0\Leftrightarrow\,\,b_{ij}\approx\infty$. Or alternatively, $a_{ij}\to 0\Leftrightarrow\,\,b_{ij}\to\infty$. Or, if the proposition is not true in general, in which cases (assumptions) would it be true?
[additional comment:]
It seems in general that is not true, but in which conditions that would be true, that is, which assumptions should we add to the problem in order to make the proposition true? Let me be a bit more specific. The matrix A is of the form $$ A=\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{bmatrix}, $$ where $A_{11}^{p\times p}$, $A_{12}^{p\times k}$, $A_{21}^{k\times p}=A_{12}^T$, $A_{22}^{k\times k}$, both $A_{11}$ and $A_{22}$ are invertible, symmetric, all the entries of A are positive. The entries of $A_{12}$, $A_{21}$ and $A_{22}$ tend to infinity. So, what happens to the entries of $A^{-1}$? Could we state at least that some of the entries tend to zero?
My feeling is that in general it is not true because the limit of many variables is not uniquely defined. In the case you multiply every $a_{ij}>0$ by the same number $t$,then you calculate the inverse and then let $t \rightarrow 0$, then it is true that every $b_{ij}$ will diverge as $1/t$, if not zero in the first place.
EDIT: I add an example, just to show what can happen for a generic parametrization $a_{ij} = a_{ij}(t)$, such that $a_{ij}(t) \rightarrow 0$ when $t \rightarrow 0$. Consider a simple $2\times 2$ case,
$$ a_{11}= t^2 \quad a_{12}=a_{21}= t \quad a_{22}= t^4 \, . $$
The inverse is (in the limit $t \rightarrow 0^+$)
$$ b_{11}= 0 \quad b_{12}= b_{21}= \infty \quad b_{22}= -1 $$
You have the whole spectrum: something is zero, something diverges, something is finite and negative.