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? :(
 A: 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 $x \to 1$, $y \to 1$.
A: What about substitution $x=s+1$, $y=t+1$?
You get
$$\frac{2xy}{x^2+y^2}-1=
\frac{2(s+1)(t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)}{(s+1)^2+(t+1)^2}-1=
\frac{2(st+s+t+1)-s^2-t^2-2s-2t-2}{(s+1)^2+(t+1)^2}=
\frac{2st-s^2-t^2}{(s+1)^2+(t+1)^2}=
\frac{-(s-t)^2}{(s+1)^2+(t+1)^2}
$$
Now you want to choose $\delta>0$ in such way that $|s|,|t|<\delta$ implies
$$\frac{(s-t)^2}{(s+1)^2+(t+1)^2} < \varepsilon.$$
If you are able to find a $\delta_1>0$ such that $|s|,|t|<\delta_1$ implies
$$(s+1)^2+(t+1)^2\ge \frac12$$
then you only need to find $\delta_2>0$ such that $|s|,|t|<\delta_2$ implies
$$(s-t)^2<\frac\varepsilon2.$$
Then you can choose $\delta=\min\{\delta_1,\delta_2\}$.
A: Hint:
With $t:=\dfrac yx$,$$f(x,y):=\frac{2xy}{x^2+y^2}=\frac{2t}{1+t^2}.$$ As this is a continuous function, it suffices to show that
$$\lim_{(x,y)\to(1,1)}\frac yx=1$$ and compute $f(1,1).$
Now you have to establish
$$\left|\frac yx-1\right|=\frac{|y-x|}{x}<\epsilon,$$ which is true when $x,y$ are sufficiently close to $1$.
