Simple multivariable limit problem

[Toshe]

Homework Statement

$f(x,y) = 2x+3y$

Let $\epsilon$ be any positive number. Show that there is a disk with center $(1,1)$ and radius $\delta$ such that whenever $P$ is in that disk, $\left| f(P) - 5\right| < \epsilon$. Give $\delta$ as a function of $\epsilon$.

Homework Equations

$\left| 2x+3y - 5\right| < \epsilon$

$\sqrt{(x-1)^2 + (y-1)^2} < \delta$

The Attempt at a Solution

Obviously, it's a continuous equation that works out to exactly $5$ so the limit is 5. But I am stuck on solving for $\delta$ in terms of $\epsilon$

I suspect I am stuck on an easy step.

 

[HallsofIvy]

$|2x+ 3y- 5|= |2x- 2+ 3y- 3|= |2(x- 1)+ 3(y- 1)|\le 2|x- 1|+ 3|y- 1|$

and, of course, $|x- 1|\le \sqrt{(x- 1)^2+ (y- 1)^2}$ and $|y- 1|\le \sqrt{(x- 1)^2+ (y- 1)^2}$.

 

[I like Serena]

Homework Statement

$f(x,y) = 2x+3y$

Let $\epsilon$ be any positive number. Show that there is a disk with center $(1,1)$ and radius $\delta$ such that whenever $P$ is in that disk, $\left| f(P) - 5\right| < \epsilon$. Give $\delta$ as a function of $\epsilon$.

Homework Equations

$\left| 2x+3y - 5\right| < \epsilon$

$\sqrt{(x-1)^2 + (y-1)^2} < \delta$

The Attempt at a Solution

Obviously, it's a continuous equation that works out to exactly $5$ so the limit is 5. But I am stuck on solving for $\delta$ in terms of $\epsilon$

I suspect I am stuck on an easy step.

Welcome to PF, Toshe!

Suppose we pick a point P with a distance r to (1,1).
Then r < δ.

What is the maximum value that |f(P)-5|=|2(x-1)+3(y-1)| can take as a function of r?