Two Dimensional Delta Epsilon Proof I was dawdling in some 2D delta epsilon examples, and I was wondering how to prove that the limit of $x^2+2xy+y^2$ has limit 3 as $(x,y)\rightarrow(1,1)$, using epsilon delta.

Let $\epsilon>0$ and we look for $\delta>0$ such that $|x^2+2xy+y^2-4|<\epsilon$ whenever $$||(x,y)-(1,1)||=\sqrt{(x-1)^2+(y-1)^2}<\delta\;(*)$$
We have
$$|x^2+2xy+y^2-4|=|(x+y)^2-4|=|(x+y-2)(x+y+2)|\le|(x+y-2)(|x|+|y|+2)|$$
Now let $\delta<1$ and with $(*)$ we have $|x|,|y|<\delta+1<2$ so
$$|(x+y-2)(|x|+|y|+2)|<6|(x+y-2)|\le6(|x-1|+|y-1|)\le12||(x,y)-(1,1)||\le12\delta$$
so it suffices to take $\delta=\min\left(\frac\epsilon{12},1\right)$ and the result follows. 
