$\pi$ -xelated

The disk has area $A_n$ and the screen has area $n^2$ . This means the percentage of the screen covered by the disk is $\frac{A_n}{n^2}$ . Now, $\lim_{n\rightarrow\infty}\frac{A_n}{n^2}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$ This is because we are essentially doing an integral: as $n$ increases, the size of each pixel becomes insignificant compared to the size of the screen, and the so the area approaches the area of a circle with a radius of a half.

Now let us look at the perimeter. Consider all the top edges of pixels that make up part of the perimeter, highlighted in red here: Since there is one line segment for every column of pixels, all the top edges total to exactly $n$ . Repeating this for the 3 other sides, we find that the perimeter is always $4n$ , so $P_n=4n$ .

This is a well known trick for falsely proving that $\pi$ equals 4. Starting with a square, we can indent the corners in such a way that the area approaches the area of a circle, but the perimeter stays the same.

Now we can find the solution to the limit in the question: $\begin{aligned}\lim_{n\rightarrow\infty}\frac{\pi_n}{\pi}&=\lim_{n\rightarrow\infty}\frac{{P_n}^2}{4\pi A_n}\\ &=\lim_{n\rightarrow\infty}\frac{16n^2}{4\pi A_n}\\ &=\frac{4}{\pi}\lim_{n\rightarrow\infty}\left(\frac{A_n}{n^2}\right)^{-1}\\ &=\frac{4}{\pi}\left(\frac{\pi}{4}\right)^{-1}\\ &=\boxed{\frac{16}{\pi^2}}\end{aligned}$