limits using $ \epsilon - \delta $ to prove two variable function I'm trying to use the $ \epsilon - \delta $ argument to prove $\lim_{(x,y) \rightarrow (1.1)}  \frac{2xy}{x^2+y^2} =1$. I know that I need to show that  $\forall  \epsilon>0,  \exists \delta>0$ s.t. for all (x,y) in the domain of f,  $| \frac{2xy}{x^2+y^2} -1| <  \epsilon$  whenever $0<  \sqrt{(x-1)^2 + (y-1)^2} <\delta$.
I've made $| \frac{2xy}{x^2+y^2} -1|$ to $\frac{(x-y)^2}{x^2 + y^2} $, but I can't get it to look like $\sqrt{(x-1)^2 + (y-1)^2}$..... what should I do from now? :(

So you have
$$
\frac{(x-y)^2}{x^2+y^2}.
$$
Since you can always assume that $x> 1/2$ and $y>1/2$, 
$$
x^2+y^2 > \frac{1}{2},
$$
and
$$
\frac{(x-y)^2}{x^2+y^2} < 2(x-y)^2.
$$
Now, $(x-y)^2 = |(x-y)(x-y)|< (|x|+|y|)|x-y| < \left(\frac{3}{2}+\frac{3}{2} \right) |x-y|$ because you can also assume $x<3/2$, $y<3/2$. I leave the verification that $|x-y|$ can be made arbitrarily small if $x \to 1$, $y \to 1$.
