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