# Delta-Epsilon Proof of a Limit with 2 Variables

1. Mar 8, 2013

### stumpoman

1. The problem statement, all variables and given/known data
Prove using the formal definition of a limit that
$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2$$
is equal to 1.
2. Relevant equations
$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2\\ \left \| \overline{x}-\overline{a} \right \|< \delta \\ \left | f(\overline{x})-L \right |<\epsilon$$

3. The attempt at a solution
$$\sqrt{(x-1)^2+(y-2)^2}<\delta\\ \left | 5x^3-x^2y^2-1 \right |<\epsilon$$
I have no idea where to go from there. I can't figure out how to manipulate the second equation to resemble the first.

2. Mar 8, 2013

### Zondrina

Perhaps I don't have the best ( biggest or smallest depending on what you want ) possible neighborhood with the delta I found, but :

$δ = min \{ 1, \sqrt[4]{ε/6} \}$ will suffice I believe.

Use the fact that $0 < |x-1|, |y-2| < δ$ to deduce it and possibly get an even better delta ( Trust me I was being lazy here, you can probably do much much better than the value I've given ).

Last edited: Mar 8, 2013
3. Mar 8, 2013

### LCKurtz

I wouldn't try to make it resemble the first. Try thinking about it this way. You do know that you can make both $|x-1|$ and $|y-2|$ small. And you are trying to make$$|5x^3-x^2y^2-1|$$small. Now, you know that $5x^3$ is going to get close to $5$, so let's subtract and add $5$:$$|5x^3-5 + 5-x^2y^2-1| = |5(x^3-1)+4-x^2y^2|$$Now that $x^2y^2$ term is going to get close to $4x^2$ as $y$ gets close to $2$, so lets add and subtract $4x^2$:$$|5(x^3-1)+4-x^2y^2 - 4x^2+4x^2|=|5(x^3-1)+4(1-x^2)+x^2(4-y^2)|$$Now do you see how to make it small?

Last edited: Mar 8, 2013
4. Mar 8, 2013

### stumpoman

so you end up with this?
$$|5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|<\epsilon$$
I see that it contains 3 terms with factors less than delta but I must be missing a some concept here.

5. Mar 8, 2013

### Zondrina

If you want to follow the path that LC is taking, apply the triangle inequality to what you have right now.

6. Mar 8, 2013

### LCKurtz

Yes, and each term has something you can make small multiplied by factors you can keep bounded.

7. Mar 11, 2013

### stumpoman

Does that mean
$$5(x^2+x+1)(x-1)-4(x-1)|x+1|-x^2(y-2)|y+2|<\epsilon$$
I don't understand.

8. Mar 11, 2013

### LCKurtz

Actually, you don't have it less than $\epsilon$ yet; that is your goal. What you do have is that$$|5x^3-x^2y^2-1|\le |5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|$$
That isn't what the triangle inequality gives you. It says $|a + b| \le |a| + |b|$ so you would get$$|5x^3-x^2y^2-1|\le |5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|$$ $$\le 5|x-1| |x^2+x+1|+4|x-1| |x+1|+|x^2||y-2||y+2|$$You want to make the whole thing less than $\epsilon$, which you could do by making each term less than $\epsilon/3$. You know you can make both $|x-1|$ and $|y-2|$ as small as you want. So if the other terms don't get too big you should be able to make it work. Remember $x$ and $y$ can't be just anything because they are getting close to $1$ and $2$, respectively. So say you make sure $\delta < 1$ so both x and y are within 1 of their limits. Then how big can the extra x and y expressions be? Overestimate some more then figure out what $\delta$ will work.