Delta-Epsilon Proof of a Limit with 2 Variables

[stumpoman]

Homework Statement

Prove using the formal definition of a limit that
$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2$$
is equal to 1.

Homework Equations

$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2\\ \left \| \overline{x}-\overline{a} \right \|< \delta \\ \left | f(\overline{x})-L \right |<\epsilon$$

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.

 

[STEMucator]

Homework Statement

Prove using the formal definition of a limit that
$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2$$
is equal to 1.

Homework Equations

$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2\\ \left \| \overline{x}-\overline{a} \right \|< \delta \\ \left | f(\overline{x})-L \right |<\epsilon$$

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.

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 ).

 

[LCKurtz]

Homework Statement

Prove using the formal definition of a limit that
$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2$$
is equal to 1.

Homework Equations

$$\lim_{(x,y) \to (1,2)} 5x^3-x^2y^2\\ \left \| \overline{x}-\overline{a} \right \|< \delta \\ \left | f(\overline{x})-L \right |<\epsilon$$

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.

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 let's 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?

 

[stumpoman]

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 let's 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?

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.

 

[STEMucator]

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.

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

 

[LCKurtz]

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.

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

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

 

[stumpoman]

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

Does that mean
$$5(x^2+x+1)(x-1)-4(x-1)|x+1|-x^2(y-2)|y+2|<\epsilon$$

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

I don't understand.

 

[LCKurtz]

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.

Actually, you don't have it less than $\epsilon$ yet; that is your goal. What you do have is that$$
|5x^3-x^2y^2-1|\le |5(x-1)(x^2+x+1)-4(x-1)(x+1)-x^2(y-2)(y+2)|$$

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

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.

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.