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?

We can choose $\delta$ such that $3x + 3y > c$ for some positive $c$, so we don't have to worry about the denominator.
For the numerator, note that:
$$|2x-y| = |2x - 2 - y + 2| \le |2x-2|+|y-2|$$
