Proving limits using $\varepsilon$-$\delta$ definition. How to verify that computed $\varepsilon$ definition is correct?

Question

So from our textbook exercises I was practicing on, it says that we need to prove this limit is true using epsilon delta definition.

$\lim_{(x,y)\to(0,2)}\ -3x +4y = 8$

I've came up with the result that $ε = δ$ , and I'm not sure if that's correct because all other examples we have had were usually $= \frac{ε}{n}$.

It would really help if I could understand how the proofs works because as far as right now I'm just copying the steps shown to us from the examples.

EDIT: so here's what I've got so far

Let $ f(x,y) = -3x +4y $ and $ L = 8 $

Let $ ε > 0 $ and take $ δ = ? $. Suppose $ 0< \sqrt{(x-0)^2 + (y-2)^2}<δ. $

$|f(x,y)- L| = |-3x+4y- 8|$ so $ |-3(x)+4(y-2)| <= -3|x| + 4|y-2|. $

since $|x| = \sqrt{(x^2)}$ <= $\sqrt{(x^2)+(y-2)^2}$ and

$|y-2| = \sqrt{((y-2)^2)}$ <= $\sqrt{(x^2)+(y-2)^2}$

so $|f(x,y)- L| = |-3x+4y- 8|$

<= $3|x| + 4|y-2|. $ <= $-3δ + 4δ = δ$

and that's how i came to the unsure conclusion that $ε = δ$.

EDIT: I understood now where I got my error from!

Answer 1

$|-3x+4(y-2)| \le 3|x|+4|y-2|$

$|x| = \sqrt{x^2} \le \sqrt{x^2+(y-2)^2}$

$|y-2| =\sqrt{(y-2)^2}\le \sqrt{x^2+(y-2)^2}$

Hence

$|-3x+4(y-2)| \le 3|x|+4|y-2| \le 7\sqrt{x^2+(y-2)^2}$

Choose $\delta < \epsilon/7$.

Then $\sqrt{x^2+(y-2)^2} < \delta$ implies

$|-3x+4(y-2)| \le 7\sqrt{x^2+(y-2)^2} \lt 7\delta \lt \epsilon$.

Answer 2

If you know the Maximum Norm (notice that Norms in $\mathbb{R}^2$ are equivalent) you can also prove it as follows:

Let be ${x \choose y} \in \mathbb{R}^2$ with $\Vert { x \choose y} - { 0 \choose 2}\Vert_{\infty} < \delta$. Without loss of generality, we assume that $\vert x - 0\vert \leq \vert y - 2 \vert < \delta$. Hence, $2-\delta< y < 2+ \delta$. Bearing this in mind we do the following:

$\vert-3x + 4y -8\vert \leq 3\vert x \vert + \vert 4y-8 \vert <3\delta + \vert 4(2+\delta)-8\vert = 3\delta +4\delta = 7\delta $.

So, if you set $\delta \leq \frac{\epsilon}{7}$ you got it.