Family of curves $x^n+y^n=a^n$ as $n$ goes from $1$ to $\infty$ (integers) and from $1$ down to $0$

Question

Take values of $n$ from 1 to $\infty$ in steps of $1$. Prove that in the limit it will be a square of side $a$. Take value of $n$ from $1$ to $0$ (fractions). In the limit as $n$ approaches $0$, prove that its a function which has value $a$ at $x=0$ and $0$ elsewhere. Any idea what happens if $n$ is negative?

Answer 1

$x^n+y^n=a^n\iff\left(\frac xa\right)^n+\left(\frac ya\right)^n=1\Longrightarrow$ If n is even, then the function $y(x)=\pm\sqrt[n]{a^n-x^n}$ can only exist for $x\in(-a,+a)$. But if n is odd, such restrictions no longer apply. Nevertheless, if we focus solely on the interval $x\in(0,a)$, then all observations that apply to even powers go for odd ones as well (i.e., the form will ultimately degenerate into a rectangle with its height twice the size of its width as $n\to\infty$).

$n\to\infty\Longrightarrow\left(\frac xa\right)^n\to0\Longrightarrow\left(\frac ya\right)^n\to1\Longrightarrow y\to a\Longrightarrow$ For $x\in(-a,a)$ we have two straight line segments characterized by $y(x)=\pm a$.

$n\to0\Longrightarrow\left(\frac xa\right)^n\to1\Longrightarrow\left(\frac ya\right)^n\to0\Longrightarrow$ y can only be $0$, since otherwise any other number risen to the power $0$ is $1$, whereas $0^0$ is indeterminate, hence $\forall x\in(0,\pm a),\quad y(x)=0$. In $x=0$, however, we have $\underset{n\to0}\lim0^n=0$, hence $y=a$.

As an aside, for $n=1$ we have a square with its four corners in $\pm a$ and $\pm ai$. For $n\to\infty$ we have another square, this time with its four corners in $a\pm ai$ and $-a\pm ai$. For $n=2$ we have the circle centered in the origin with the radius $a$.