A big LED screen

For $n > 8$ , the LEDs that are not in the first $4$ columns, first $4$ rows, last $4$ columns, and last $4$ rows will be toggled $25$ times (one for each position in the $5 \times 5$ square) and therefore will be on at the end.

Therefore, at least the $(n - 8) \times (n - 8) = n^2 - 16n + 64$ central square of LEDs will be on at the end, and at most all $n \times n = n^2$ LEDs will be on at the end.

In other words, $\displaystyle \lim_{n \to \infty} \frac{n^2 - 16n + 64}{n^2} \leq \lim_{n \to \infty} \frac{\text{on}(n)}{n^2} \leq \lim_{n \to \infty} \frac{n^2}{n^2}$ .

Since $\displaystyle \lim_{n \to \infty} \frac{n^2 - 16n + 64}{n^2} = 1$ and $\displaystyle \lim_{n \to \infty} \frac{n^2}{n^2} = 1$ , by the squeeze theorem $\displaystyle \lim_{n \to \infty} \frac{\text{on}(n)}{n^2} = \boxed{1}$ as well.

For large $n$ , most of the LEDs will be toggled 25 times and will eventually be ON. The only ones that will be toggled an even number of times are some of those near the edges and corners. It doesn't matter exactly how many, though, because as we go to infinity, they will be overwhelmed by the internal ones. Almost all of the LEDs will be on as the proportion goes to $\boxed{1}$ .