Prove that $\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$ How to prove that $$\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$$ with $\varepsilon-\delta$ limit definition?
I know so far that $\forall \varepsilon > 0 \  \exists \delta > 0: 0 < d((x, y), (1, 2)) < \delta \Rightarrow \Big|\frac{x}{x+y} - \frac{1}{3}\Big| < \varepsilon$, so we get
$d = \sqrt{|x-1|^2 + |y-2|^2} \Rightarrow d^2 = |x-1|^2 + |y-2|^2 \Rightarrow \begin{cases}
             |x - 1| < d < \delta\\
             |y - 2| < d < \delta\\
           \end{cases}$
But I don't know how to express $\Big|\frac{x}{x+y} - \frac{1}{3}\Big| = \Big|\frac{3x-(x+y)}{3(x+y)}\Big| = \Big|\frac{3x-x-y}{3x+3y}\Big| = \Big|\frac{2x-y}{3x+3y}\Big|$ using $|x-1|$ and $|y-2|$.
Can anyone help me with that?

By $|x-1|<\delta$ and $|y-2|<\delta$ assuming wlog $\delta\le\frac12$ we have
$$\left|\frac{x}{x+y} - \frac{1}{3}\right|=\left|\frac{2x-y}{x+y}\right|=\left|\frac{2(x-1)-(y-2)}{x+y}\right| \le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right|\le$$
$$\le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right| \le|x-1|+\frac12|y-2|=\frac32 \delta$$
