Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$ 
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$

Take $\epsilon>0$, I want to find $\delta>0$ such that:
$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert<\epsilon$$
So I started by adding both fractions and obtained:
$$\left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y-2+2}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2(x-1)-(y-2)}{x+y}\right\rvert$$
Now, I have $\lvert x-1\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$ and $\lvert y-2\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$
However, im not being able to bound $\frac{1}{\lvert x+y\rvert}$
Am I on the correct track? Any suggestions? Thank you very much. 

On bounding $\frac{1}{|x+y|}$. If you take $\delta<1/2$, then 
$$
|x-1|<\delta
$$
and 
$$
|y-2|<\delta 
$$
yields 
$$
1/2<x
$$
and
$$
3/2<y
$$
and so 
$$
|x+y|=x+y>2
$$
which is the part you are worried about.
The intuition behind this: Note that you are worried about $|x+y|$ getting very small, but of course it wont since in the limit, $x\approx 1$ and $y\approx 2$.
