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

Question

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.

Answer 1

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 $y\approx 2$.

Answer 2

$\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 $.