Question:
For $t\in\mathbb{N}^+$, the limit Set-theoretic limit of the region$$\mathcal{R}_t=\{(x,y)\in\mathbb{R}^2\mid x^{2t-1}(x-y)(x-1)+y^{2t-1}(y-1)(y-x)+(1-x)(1-y)\ge0\}$$ as $t\to\infty$ is the region $\mathcal{R}_\infty=\{(x,y)\in\mathbb{R}^2\mid\max(1-\max(|x|,|y|),x+y)\ge0\vee y=x\}$.
Attempt:
$\{(x,y)\in\mathbb{R}^2\mid\max(1-\max(|x|,|y|),x+y)\ge0\vee y=x\}$ is the union of the regions $S:=\{(x,y)\in\mathbb{R}^2\mid\max(|x|,|y|)\le1\}$ and $H:=\{(x,y)\in\mathbb{R}^2\mid x+y\ge0\}$ and the line $y=x$.
The line $y=x$ is contained in all $\mathcal{R}_t$, so it is contained in $\mathcal{R}_\infty$.
$S$ is the square with vertices $(1,1),(1,-1),(-1,-1),(-1,1)$, and $H$ is the half-plane bounded by the line $x+y=0$.
For all $(x,y)\in S$, we have $|x|\le1$ and $|y|\le1$, so $x^{2t-1}\to0$ and $y^{2t-1}\to0$ as $t\to\infty$. Thus, the limit of the left-hand side of the inequality in the question is $(1-x)(1-y)$, which is nonnegative. Therefore, $S$ is contained in $\mathcal{R}_\infty$.
For all $(x,y)\in H$, we have $x+y\ge0$, so $x^{2t-1}+y^{2t-1}\ge0$, but the sign of the other terms is not clear. I am not sure how to proceed from here. Any help would be appreciated.
