# Homework Help: Epsilon-Delta Proof with 2-Variable Polynomial

1. Sep 23, 2013

### idochemistry

1. The problem statement, all variables and given/known data
Find a specific number δ>0 such that if x2 + y2 = δ2, then |x2+y2+3xy+180xy5 < 1/10 000.

2. Relevant 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.

3. 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.

2. Sep 23, 2013

### jbunniii

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 $\delta$?

3. Sep 23, 2013

### 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|)

4. Sep 23, 2013

### jbunniii

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 $\sqrt{x^2 + y^2}$?

5. Sep 23, 2013

### Staff: Mentor

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.

6. Sep 23, 2013

### idochemistry

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!

7. Sep 23, 2013

### jbunniii

So far so good.

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}

8. Sep 23, 2013

### 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?

9. Sep 23, 2013

### jbunniii

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.

10. Sep 23, 2013

### jbunniii

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.

11. Sep 23, 2013

### idochemistry

Waiiiiit. 4δ2 + 180δ6

12. Sep 23, 2013

### idochemistry

would be the right side?

13. Sep 23, 2013

### jbunniii

Yes, that looks right:

$x^2 + y^2 < \delta^2$
$3|x||y| < 3\delta^2$
$180|x||y|^5 < 180\delta^6$

so

$x^2 + y^2 + 3|x||y| + 180|x||y|^5 < 4\delta^2 + 180\delta^6$