Can we graph $x^{\infty}-y^{\infty}=1$

Question

Playing around with exponents on $x^{n}-y^{n}=1$ suggests a trend toward cleaner and cleaner corners as n increases on one graph for even numbers and a second graph for odd numbers. Therefore, is it possible to graph the case of $n=\infty$ for odd and/or even numbers (separately)?

Answer 1

Odd case:

For every $k$, $y_{2k+1}(x)=(x^{2k+1}-1)^{1/(2k+1)}$ hence, when $k\to\infty$, $y_{2k+1}(x)\to x$ if $x\leqslant-1$, $y_{2k+1}(x)\to-1$ if $-1\leqslant x\lt1$, $y_{2k+1}(1)=0$, and $y_{2k+1}(x)\to x$ if $x\gt1$.

Even case:

For every $k$, $y_{2k}(x)=\pm(x^{2k}-1)^{1/(2k)}$ hence, when $k\to\infty$, $y_{2k}(x)$ does not exist if $|x|\lt1$, $y_{2k}(x)=0$ if $|x|=1$ and $y_{2k}(x)\to\pm x$ if $|x|\gt1$.

Proofs:

Use only the fact that $|x|^n\to\infty$ if $|x|\gt1$ and $|x|^n\to0$ if $|x|\lt1$.

Answer 2

It does not make sense to use $\infty$ in a function. You may however have a limiting curve for either case.