<s>[INST] 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. 
 [/INST] $\lim_{(x,y) \to (1,2)}\dfrac{3x-4y}{x+y}
=-\dfrac{5}{3}
$
I like to let
variables go to zero,
so let
$x = 1+u, y=2+v$.
Then
$\begin{array}\\
\dfrac{3x-4y}{x+y}+\dfrac{5}{3}
&=\dfrac{3(1+u)-4(2+v)}{1+u+2+v}+\dfrac{5}{3}\\
&=\dfrac{3u-4v-5}{3+u+v}+\dfrac{5}{3}
\qquad\text{You can see here that the limit will be -5/3}\\
&=\dfrac{3(3u-4v-5)+5(3+u+v)}{3(3+u+v)}\\
&=\dfrac{14u-7v}{3(3+u+v)}
\qquad\text{and the constant term cancels out}\\
&=\dfrac{14u-7v}{9+3(u+v)}\\
\end{array}
$
From this we see that
if $u$ and $v$ are small
the the numerator is small
and the denominator
is not small,
being around $9$.
To make this rigorous,
if $|u|, |v| < c$
where
(to get a lower bound
on the denominator)
$0 < c < \frac12$,
then
$|14u-7v|
\lt 21c
$ and
$|9+3(u+v)|
\ge 9-3|u+v|
\ge 9-3(2c)
\gt 6
$
so
$|\dfrac{14u-7v}{9+3(u+v)}|
\le \dfrac{21c}{6}
=\dfrac{7c}{2}
$.
Therefore,
to make
$|\dfrac{14u-7v}{9+3(u+v)}|
\lt d$,
choose
$\dfrac{7c}{2}
\lt d$
or
$c \lt \dfrac{2d}{7}
$.
Finally,
to go back to
the original problem,
if
$|x-1| < \dfrac{2d}{7}$
and
$|y-2| < \dfrac{2d}{7}$
where
$\dfrac{2d}{7} < \frac12$
then
$|\dfrac{3x-4y}{x+y}+\dfrac{5}{3}|
\lt d
$.
</s>