Q: Passing to a circle function to a square function Consider the equation $x^2+y^2=1$. It will describe a circle. Hand to hand we increase the exponent with EVEN numbers, the circle begins becoming a square. So we'll arrive to $x^∞ + y^∞ =1$ and the circle will become a square. The circle will be inscribed in the square, like you can see here https://www.desmos.com/calculator/hhmvadmjc9. How can I demonstrate this passage? How can I take over that change of the function? And why does the second function describe a square?
 A: The notation $x^\infty$ and $y^\infty$ means nothing. The best point of view is norm.
The classical euclidean norm $\|\cdot \|_2$ is defined by ($\overrightarrow x=(x_1,x_2)$) 
$$\| \overrightarrow x \|_2 = \sqrt{x_1^2+x_2^2}$$
You can check that the set of $\overrightarrow x$ such that $\| \overrightarrow x \|_2 = 1$ is exactly the standard unit circle.
At the opposite, you can define another norm $\|\cdot \|_\infty$  by
$$\| \overrightarrow x \|_\infty = \max(\vert x_1 \vert,\vert x_2 \vert)$$
and you can check that the set of $\overrightarrow x$ such that $\| \overrightarrow x \|_\infty = 1$ is your square.
You can also define intermediate norms $\|\cdot \|_p$  by
$$\| \overrightarrow x \|_p = (x_1^p+x_2^p)^{1/p}$$
The set of $\overrightarrow x$ such that $\| \overrightarrow x \|_p = 1$ looks like rounded square sharper the more $p$ is large.
Actually one can show that
$$\| \overrightarrow x \|_p \underset{p \to +\infty}{\longrightarrow} \| \overrightarrow x \|_\infty.$$
This justify your affirmation.
