How to prove $\lim \limits_{(x,y) \to (3,5)}(2x-5y)=-19$ using $\epsilon - \delta$ definition

Question

I am learning the $\epsilon$ - $\delta$ definition of a limit and I am struggling to prove the following limit:

$\lim \limits_{(x,y) \to (3,5)}(2x-5y)=-19$

I know where to start, which would be stating that for every $\delta$ > 0 where $0< \sqrt{(x-3)^2+(y-5)^2} < \delta$, there exists some $\epsilon$ such that $|2x-5y-(-19)| < \epsilon$.

I am stuck with the actual work of the limit, however.

I started out by stating that $|2x-5y+19|\leq |2x| +|5y| +19$ but I don't know how to continue. Any ideas on how to solve this would be greatly appreciated.

Answer 1

It is always easier to work in zero for limits than at a given number.

So just set $x=3+u$ and $y=5+v$ with $(u,v)\to 0$.

Then you get $$\Big|2x-5y+19\Big|=\Big|2(3+u)-5(5+v)+19\Big|=\Big|2u-5v\Big|\le 2|u|+5|v|\to 0$$

Remark: the norm chosen in $\mathbb R^2$ does not matter as, all norms are equivalent in finite dimension. You can have a look at my post here https://math.stackexchange.com/a/3916803/399263

Anyway would you need to use the euclidean norm you can proceed like this:

$2|u|+5|v|\le 5(|u|+|v|)=5\lVert(u,v)\lVert_1\le 10\lVert(u,v)\lVert_2\to 0$

Answer 2

$f(x,y):=$

$|2(x-3)+6-5(y-5)¬25+19|$

$=|2(x-3)-5(y-5)|\le$

$2|x-3|+5|y-5|;$

Choose $\delta=\epsilon/7$;

Then

$f(x,y)\le 2|x-3|+5|y-5| <$

$2\delta +5\delta=7\delta <\epsilon$.

Note:

$\sqrt{(x-3)^2+(y-5)^2}<\delta$ implies

$|x-3|<\delta$ and $|y-5|<\delta$.