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 $\epsilon&gt;0$, I want to find $\delta&gt;0$ such that:

$$\lVert (x-1,y-2)\rVert &lt;\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert&lt;\epsilon$$

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 $\lvert x-1\rvert\leq\sqrt{(x-1)^2+(y-2)^2}&lt;\delta$ and $\lvert y-2\rvert\leq\sqrt{(x-1)^2+(y-2)^2}&lt;\delta$

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.
$\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}
&amp;=\dfrac{3(1+u)-4(2+v)}{1+u+2+v}+\dfrac{5}{3}\\
&amp;=\dfrac{3u-4v-5}{3+u+v}+\dfrac{5}{3}
\qquad\text{You can see here that the limit will be -5/3}\\
&amp;=\dfrac{3(3u-4v-5)+5(3+u+v)}{3(3+u+v)}\\
&amp;=\dfrac{14u-7v}{3(3+u+v)}
\qquad\text{and the constant term cancels out}\\
&amp;=\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 $|u|, |v| &lt; c$
where
(to get a lower bound
on the denominator)
$0 &lt; c &lt; \frac12$,
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
$|x-1| &lt; \dfrac{2d}{7}$
and
$|y-2| &lt; \dfrac{2d}{7}$
where
$\dfrac{2d}{7} &lt; \frac12$
then
$|\dfrac{3x-4y}{x+y}+\dfrac{5}{3}|
\lt d
$.
On bounding $\frac{1}{|x+y|}$. If you take $\delta&lt;1/2$, then 
$$
|x-1|&lt;\delta
$$
and 
$$
|y-2|&lt;\delta 
$$
yields 
$$
1/2&lt;x
$$
and
$$
3/2&lt;y
$$
and so 
$$
|x+y|=x+y&gt;2
$$
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 $y\approx 2$.