# Two dimensional limit proof

1. Jul 13, 2012

### Zondrina

1. The problem statement, all variables and given/known data
This is my first attempt at a multivariate ε-δ proof. I don't know if I'm doing it right or wrong so I'm hoping someone can shed some light.

Prove :
$$\lim_{(x,y)\rightarrow(1,1)} x^2+y^2 = 2$$

2. Relevant equations
$\forall$ε>0, $\exists$δ>0 | 0<|x-1||y-1|<δ $\Rightarrow$ |x^2+y^2-2|<ε

3. The attempt at a solution

So we know : |x-1|<δ and |y-1|<δ

|x^2+y^2-2| = |x^2-1+y^2-1| ≤ |x^2-1|+|y^2-1| = |x+1||x-1| + |y+1||y-1| < δ|x+1| + δ|y+1|

Now let us express |x+1| and |y+1| in terms of δ so we can further simplify to our desired value of δ.

We can observe that : |x+1| < δ+2 and |y+1| < δ+2

So we continue forth : δ|x+1| + δ|y+1| < δ(δ+2) + δ(δ+2)

Now for convenience let us agree to allow δ to be bounded, say δ≤1 for simplicity.

Now we get : δ(δ+2) + δ(δ+2) ≤ 6δ ≤ ε

Therefore δ=min{1,ε/6}

So now we verify δ satisfies the definition.

$\forall$ε>0, $\exists$δ=min{1,ε/6}>0 | 0<|x-1||y-1|<δ $\Rightarrow$ |x^2+y^2-2|<ε

|x^2+y^2-2| = |x^2-1+y^2-1| ≤ |x^2-1|+|y^2-1| = |x+1||x-1| + |y+1||y-1| < δ|x+1| + δ|y+1|| < δ(δ+2) + δ(δ+2) ≤ 3ε/6 + 3ε/6 = ε

Thus δ=min{1,ε/6} $\Rightarrow$ |f(x,y)-L|<ε and the proof is concluded. I want to know if I'm doing this correctly or if I've missed something.

2. Jul 14, 2012

### SammyS

Staff Emeritus
When you state
$\forall$ε>0, $\exists$δ>0 | 0<|x-1||y-1|<δ $\Rightarrow |x^2+y^2-2|<ε$​

You need to say something like 0<|x-1|<δ and 0<|y-1|<δ instead of 0<|x-1||y-1|<δ .

If 0<|x-1||y-1|<δ, and |x-1| is very small, then |y-1| could be many times larger than δ.

Your last three lines constitute the actual proof. The part above that shows how you find the δ you need.

In my opinion, you need to be more explicit regarding how your value for δ makes $|x^2+y^2-2|<\varepsilon\ .$

.

3. Jul 14, 2012

### HallsofIvy

Staff Emeritus
Strictly speaking you should have $0< \sqrt{(x-1)^2+ (y-1)^2}< \delta$ but it can be shown that that is equivalent to $0< |x-1|<\delta$ and $0<|y- 1|<\delta$. Another equivalent form is $0< |x-1|+ |y- 1|$.

But the product, $0< |x-1||y-1|<\delta$, is NOT acceptable. I suspect that what you are remembering was "$0< |x-1|,|y-1|<\delta$" where the comma indicates that these are two separate inequalities that have been combined for brevity.

Last edited: Jul 14, 2012
4. Jul 14, 2012

### Zondrina

Ah i see what you mean. I suppose I'll start using a comma to be more precise about what i intend. Just one thing, is your |x--1| a typo?

Also @ sammy, how would i be more explicit? Would i actually mention when and where i was substituting my δ when i was proving that it worked?

5. Jul 15, 2012

### SammyS

Staff Emeritus
Maybe I should have said 'more direct'.

If δ = min(ε/6 , 1), then
|x-1| < 1 ➾ x-1 < 1 ➾ |x+1| < 3

and |x-1| < ε/6

Therefore, |x+1||x-1| < (3)(ε/6) = ε/2 ​

A similar argument holds for y, so that |y+1||y-1| < ε/2 .

...

6. Jul 15, 2012

### Zondrina

I understand what you mean. Where would I mention this though? I'm guessing when I got to this point I would do this :

If δ=min{1,ε/6} then we observe :

∀ε>0, ∃δ=min{1,ε/6}>0 | 0<|x-1|,|y-1|<δ ⇒ |x^2+y^2-2|<ε

So we know :
|x-1|<1 $\Rightarrow$ |x+1|<3 and also |x-1|< ε/6
|y-1|<1 $\Rightarrow$ |y+1|<3 and also |y-1|< ε/6

So we have : |x^2+y^2-2| = |x^2-1+y^2-1| ≤ |x^2-1|+|y^2-1| = |x+1||x-1| + |y+1||y-1| < (3)(ε/6)+(3)(ε/6) = ε

and that would be how I would want to style it ( Skipping the δ portion and going straight to the end like that ). Thanks for the help btw, style is super important to me so I'm glad someone knows what they're doing to help me :)