# Thread: Multivariable Delta Epsilon Proof

1. ## Multivariable Delta Epsilon Proof

Q: limit of f(x,y)=(x^3-y^3)/(x^2+y^2) as (x,y) -> (0,)?

What I tried:
1. Picked 0 as L
2. We know x^2+y^2 < d^2
3. and lxl=<d and lyl=<d

I can't think of an inequation needed

Any help would be greatly appreciated. Thanks

2. Originally Posted by 6DOM
Q: limit of f(x,y)=(x^3-y^3)/(x^2+y^2) as (x,y) -> (0,)?

What I tried:
1. Picked 0 as L
2. We know x^2+y^2 < d^2
3. and lxl=<d and lyl=<d

I can't think of an inequation needed

Any help would be greatly appreciated. Thanks
I think you mean inequality. As

$\displaystyle - \left(x^2+y^2\right)^{1/2} \le x,y \le \left(x^2+y^2\right)^{1/2}$

try

$\displaystyle - \left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2}.$

3. Originally Posted by Danny
I think you mean inequality. As

$\displaystyle - \left(x^2+y^2\right)^{1/2} \le x,y \le \left(x^2+y^2\right)^{1/2}$

try

$\displaystyle - \left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2}.$
I'm not sure I understand you.

I thought we needed some kind of inequality that looks like

l(x^3-y^3)/(x^2+y^2)l =<

so that we can set delta

4. Originally Posted by 6DOM
I'm not sure I understand you.

I thought we needed some kind of inequality that looks like

l(x^3-y^3)/(x^2+y^2)l =<

so that we can set delta
If

$\displaystyle \left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2},$

then

$\displaystyle \left(x^2+y^2\right)^{1/2} \le \frac{x^3 - y^3}{x^2+y^2} \le \left(x^2+y^2\right)^{1/2},$

then

$\displaystyle \left|\, \frac{x^3 - y^3}{x^2+y^2} \, \right| \le \left(x^2+y^2\right)^{1/2}.$