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.
 A: Using Cauchy-Schwarz inequality twice: 
$(x^2 + 2xy + y^2 - 4)^2 = ((x+y)^2 - 2^2)^2 = (((x-1) + (y-1))((x-1) + (y-1) + 4))^2$ <= 
$2*((x-1)^2 + (y-1)^2)*18((x-1)^2 + (y-1)^2 + 1) = S$ . 
So let epsilon e > 0 be given, we need to choose delta = d > 0 so that the square-root of S is less than epsilon e. First we want:
$(x-1)^2 + (y-1)^2 < 3$
so we can take square-root and get a whole number. So we force d to be less than 3^(1/2). So we also want that $12*((x-1)^2 + (y-1)^2)^(1/2) < e$
==>
$((x-1)^2 + (y-1)^2)$^(1/2) < e/12. So naturally, we choose 
delta d = min { e/12, 3^(1/2) } > 0.
A: 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. 
