# Homework Help: Epsilon-Delta Limit in Two Variables

1. Oct 23, 2012

### Vorde

1. The problem statement, all variables and given/known data

Show (using the epsilon-delta method) that the $\lim_{\substack{x\rightarrow a\\y\rightarrow b}} y = b$

2. Relevant equations

Epsilon-Delta Definition

$\lim_{\substack{x\rightarrow a\\y\rightarrow b}} f(x,y) = L$
means that for every $\epsilon$ > 0, there exists a number $\delta$>0 such that:
$|f(x,y)-L| < \epsilon$ whenever $0 < \sqrt{(x-a)^2+(y-a)^2} < \delta$

3. The attempt at a solution

I totally get what the epsilon-delta definition is and what it means, but I'm not so sure how to go about showing this limit exists.
My textbook has only one example and it's really badly described.

My understanding is that I have
$|y-b| < \epsilon$ when $\sqrt{x^2+y^2} < \delta$
So I can say $y≤\sqrt{x^2+y^2}$
So $|y-b| ≤ \sqrt{x^2+y^2}$?

But I don't get where to go from here. What should my endgoal be?

2. Oct 23, 2012

### Robert1986

So, if $\sqrt{(x-a)^2 + (y-b)^2} < \delta$, what can you say about how big $\sqrt{(y-b)^2}$ is?

3. Oct 23, 2012

### Vorde

$\sqrt{(y-b)^2} < \sqrt{(y-a)^2+(x-a)^2}$
So
$\sqrt{(y-b)^2} < \delta$
But I don't understand what I'm supposed to show.
Is it enough to say what I just said? I feel like I need more...

Last edited: Oct 23, 2012
4. Oct 23, 2012

### Robert1986

Well, so now you need to pick what your $\delta$ is, right? So, $|y-b| = \sqrt{(y-b)^2} < \delta$, so what should you pick your $\delta$ to be?

5. Oct 23, 2012

### Zondrina

You know what you want, so apply the definition to show what you want.

$\forall ε>0, \exists δ > 0 \space | \space 0 < |x-a|, |y-b| < δ \Rightarrow |y - b| < ε$

Now if |y-b| < ε then we are already done. Now choose any arbitrary δ ≤ aε where $a \in ℝ, 0 < a < 1 \space$ as to make the neighborhood around f(x,y) as small or large as we like and still satisfy the definition.

Really depends on how big you want your neighborhood to be right :) ?

6. Oct 23, 2012

### Vorde

Ah! I got it!

Thanks to all of you :)
One less problem left in the world.