$\lim_{x\to\infty} K_{ij}(x) = 0$ BUT $\lim_{x\to\infty} ||K(x)||_{F} = \sqrt{n-1}$ !!!

Question

I have a real square full-rank matrix $K$ of size $n\times n$, where each entry of this matrix is a function of a real variable $x$. As I compute numerically the limit as $x \to \infty$, every entry goes to zero, which is what I expect, knowing pretty well the form of my matrix entries (essentially, these are all exponentials of negative arguments). However, and this is what I find mindblowing, taking the limit of the Frobenius norm of my matrix, what I get is not $0$, as I'd expect, but exactly $\sqrt{n-1}$. How's that even possible or is there just some mistake in my code ?!!

Answer 1

Since $$\|K(x)\|_F^2 = \sum_{i,j=1}^{n}K_{i,j}(x)^2,$$ we have $$\lim_{x \to \infty}\|K(x)\|_F^2 = \lim_{x \to \infty}\sum_{i,j=1}^{n}K_{i,j}(x)^2 = \sum_{i,j=1}^{n}\lim_{x \to \infty}K_{i,j}(x)^2 = \sum_{i,j=1}^{n}0 = 0,$$ where the interchange of the limit and summation is valid since the sum has finitely many terms. So what you are describing is impossible.

If you were taking the limit as $n \to \infty$ instead of $x \to \infty$, i.e., the matrix was growing in size, then it would be possible to have each entry of the matrix tend to $0$, but the Frobenius norm grow.