Let $A\in R^{n\times n}$ (symmetric), I know that, $\forall(i,j)$ $\lim_{x\to\infty} a_{ij}(x)=0$, .
I want to show, if possible, that the $$\lim_{x\to\infty} A^{-1}=E,$$ where $E$ is a $n\times n$ matrix with all the entries equal to $\infty$.
We know that the inverse is a continuous map, we have that $\lim_{x\to\infty} A^{-1}=(\lim_{x\to\infty} A)^{-1}$. Any ideas of how to follow?
This is not right. Consider the matrices $$ A_k = \begin{pmatrix} 1/k & 1/k^2 \\ 1/k^2 & 1/k \end{pmatrix}. $$ Then $$ A_k^{-1} = \frac{k^2}{k^2-1} \begin{pmatrix} k & -1 \\ -1 & k \end{pmatrix} \to \begin{pmatrix} + \infty & -1 \\ -1 & + \infty \end{pmatrix}. $$