# Epsilon-Delta Proof with 2-Variable Polynomial

idochemistry

## Homework Statement

Find a specific number δ>0 such that if x2 + y2 = δ2, then |x2+y2+3xy+180xy5 < 1/10 000.

## Homework Equations

ε-δ def'n of limit: lim (x,y) → (a, b) f(x) = L if for every ε > 0 there exists a δ > 0 such that 0 < √(x-a)2+(y-b)2, |f(x) - L| < ε.

## The Attempt at a Solution

So, f(x) = |x2+y2+3xy+180xy5 and because x2 + y2 = δ2, the limit is being taken as (x, y) → (0, 0) and L = 0.
I don't understand how I would start this problem because it's not rational, like I'm used to. I can do this:
(x) = |x2+y2+3xy+180xy5 = x2 + y2 + 3|x||y| + 180|x||y5|, but now I am stuck. Any help is greatly appreciated.

Homework Helper
Gold Member
|x2+y2+3xy+180xy5| = x2 + y2 + 3|x||y| + 180|x||y5|, but now I am stuck. Any help is greatly appreciated.
Well, this equation is certainly not true, but if you replace the ##=## with ##\leq##, then it is true, because of the triangle inequality. So this gives you
$$|x^2 + y^2 + 3xy + 180xy^5| \leq |x^2 + y^2| + |3xy| + |180xy^5|$$
Assuming ##x## and ##y## are real, it's true that ##|x^2 + y^2| = x^2 + y^2##. And the absolute value of a product equals the product of absolute values. So we can rewrite the right hand side as
$$x^2 + y^2 + 3|x|\cdot |y| + 180 |x| \cdot |y|^5$$
You know that ##x^2 + y^2 = \delta^2##. Can you find a way to bound the other terms in a way that depends on ##\delta##?

idochemistry
Sorry, I meant to say that x2+y2 < δ^2

I am still very confused.

Can I do something like this?

0 ≤|x2 + y2 + 180xy5| ≤ x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5| = |3x2y2(|1+60y3|)

Homework Helper
Gold Member
0 ≤|x2 + y2 + 180xy5| ≤ x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5| = |3x2y2(|1+60y3|)
It isn't necessarily true that ##|3xy| \leq |3x^2 y^2|## or that ##|180xy^5| \leq |180x^2 y^5|##. This is true if ##x \geq 1## and ##y \geq 1## but not in general.

Here is a hint. How does ##|x|## compare to ##\sqrt{x^2 + y^2}##?

Mentor
Sorry, I meant to say that x2+y2 < δ^2

I am still very confused.

Can I do something like this?

0 ≤|x2 + y2 + 180xy5| ≤ x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5| = |3x2y2(|1+60y3|)

No. You can't turn a sum into a product like that.

IOW, you can't do this: x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5|

And you can't do what you did in the last expression, either.

idochemistry
Here is a hint. How does ##|x|## compare to ##\sqrt{x^2 + y^2}##?

Okay. So,
|x| ≤ $\sqrt{}x2+y2$< δ
and
|y| ≤ $\sqrt{}x2+y2$< δ

Then,
|x2+y2+3xy+180xy5| < δ+δ+3δ2+180δ6

Right?

Also, thank you so much!

Homework Helper
Gold Member
Okay. So,
|x| ≤ $\sqrt{}x2+y2$< δ
and
|y| ≤ $\sqrt{}x2+y2$< δ
So far so good.

Then,
|x2+y2+3xy+180xy5| < δ+δ+3δ2+180δ6
That's not quite right. Try writing out the intermediate steps:
\begin{align} |x^2 + y^2 + 3xy + 180xy^5| & \leq x^2 + y^2 + 3|x| |y| + 180 |x| |y|^5 \\ & \leq \delta^2 + 3|x| |y| + 180 |x| |y|^5 \\ &\leq ???\end{align}

idochemistry
Isn't |x2+2+3xy+180xy5| ≤ δ2
ie. it doesn't matter what you add to the right side because it will always make δ2 larger, in this case?

Homework Helper
Gold Member
You're trying to solve for ##\delta## which will satisfy the requirement in your problem statement. You need to find a bound for ##|x^2 + y^2 + 3xy + 180xy^5|## that depends only on ##\delta##, not ##x## or ##y##. You have the constraint ##x^2 + y^2 < \delta^2##. From this, in post #6 you found some bounds for ##|x|## and ##|y|##. Now apply those to the right hand side in post #7.

Homework Helper
Gold Member
Isn't |x2+2+3xy+180xy5| ≤ δ2
ie. it doesn't matter what you add to the right side because it will always make δ2 larger, in this case?
To address this question directly, no, it's not necessarily true. All you know is that ##x^2 + y^2 < \delta^2##. But ##|x^2 + y^2 + 3xy + 180xy^5|## may be bigger than ##x^2 + y^2##, so it isn't necessarily bounded by ##\delta^2##. Try plugging in ##x=y=1## and ##\delta = 2## to see this.

idochemistry
Waiiiiit. 4δ2 + 180δ6

idochemistry
would be the right side?