Homework Help: Solve this multivariable limit using epsilon-delta methods

MeMoses · Sep 25, 2012

1. The problem statement, all variables and given/known data

Compute the following limit with the epsilon-delta method
lim(x,y)->(0,0) of (x**3-y**3) / (x**2+y**2)

2. Relevant equations

3. The attempt at a solution
I don't remember much about the epsilon-delta method and I haven't used it for multivariable limits. I tried abs( f(x) ) <= abs(x**3)/(x**2+y**2) - abs(y**3)/(x**2+y**2) and then abs(x**3)/(x**2+y**2) <= abs(x**3)/x**2 = abs(x) and I did the same for y. Am I even going in the right direction and if so how would I finish this up?

SammyS · Sep 25, 2012

MeMoses said: ↑

1. The problem statement, all variables and given/known data

Compute the following limit with the epsilon-delta method
lim(x,y)->(0,0) of (x**3-y**3) / (x**2+y**2)

2. Relevant equations

3. The attempt at a solution
I don't remember much about the epsilon-delta method and I haven't used it for multivariable limits. I tried abs( f(x) ) <= abs(x**3)/(x**2+y**2) - abs(y**3)/(x**2+y**2) and then abs(x**3)/(x**2+y**2) ,= abs(x**3)/x**2 = abs(x) and I did the same for y. Am I even going in the right direction and if so how would I finish this up?

Expand the difference of cubes.

x³ - y³ = (x - y)(x² + xy + y²)

Change to polar coordinates.

MeMoses · Sep 25, 2012

How could I go about it without the use of polar coordinates? I would like to see the different ways to solve this.

Zondrina · Sep 25, 2012

MeMoses said: ↑

How could I go about it without the use of polar coordinates? I would like to see the different ways to solve this.

Well, you know that |x| < δ and |y| < δ from the definition... use this to your advantage.

micromass · Sep 25, 2012

I would think of the squeeze theorem.

MeMoses · Sep 25, 2012

if I use |x| < δ and |y| < δ from the work I did initially the δ just cancel out and I get |f(x)| < δ-δ=0 which I know can't be right.

Zondrina · Sep 25, 2012

MeMoses said: ↑

if I use |x| < δ and |y| < δ from the work I did initially the δ just cancel out and I get |f(x)| < δ-δ=0 which I know can't be right.

Think about what you're saying here...

Remember also that |x| ≤ (x^2+y^2)^(1/2) in this scenario...

MeMoses · Sep 25, 2012

Well now that I look at it, I cannot split up the absolute value in the numerator like I did. For instance if x=-2 and y=1, I would get(in the numerators) 9<=8-1 which is not true. So what other ways can I approach this?

Zondrina · Sep 25, 2012

MeMoses said: ↑

Well now that I look at it, I cannot split up the absolute value in the numerator like I did. For instance if x=-2 and y=1, I would get(in the numerators) 9<=8-1 which is not true. So what other ways can I approach this?

Apply the definition.

[itex]\forall ε>0, \exists δ>0 \space | \space 0<|(x,y)-(0,0)|<δ \Rightarrow |f(x,y)-L|<ε[/itex]

Now massage the |f(x,y)-L| and use the fact that (x^2+y^2)^(1/2) < δ to finish your proof.

MeMoses · Sep 25, 2012

Ok heres what I got. Correct me if I'm wrong.
If I fix my mistake I can set |f(x)| <= |x^3|/(x^2 + y^2) + |y^3|/(x^2+y^2) <= |x| + |y| < δ + δ = epsilon. Thus |f(x)-0| < 2δ=epsilon => lim = 0. I think that all makes sense. When you said massage the |f(x,y)-L| did you mean something like this or is there a better way for me to go about these problems, because I am not seeing it.

Zondrina · Sep 25, 2012

MeMoses said: ↑

Ok heres what I got. Correct me if I'm wrong.
If I fix my mistake I can set |f(x)| <= |x^3|/(x^2 + y^2) + |y^3|/(x^2+y^2) <= |x| + |y| < δ + δ = epsilon. Thus |f(x)-0| < 2δ=epsilon => lim = 0. I think that all makes sense. When you said massage the |f(x,y)-L| did you mean something like this or is there a better way for me to go about these problems, because I am not seeing it.

[itex]|f(x,y)-L| = |\frac{x^3-y^3}{x^2+y^2} - 0| = \frac{|x^3-y^3|}{|x^2+y^2|}[/itex]

From there you would massage your expression by using things like the triangle inequality and any useful facts you know into the form (x^2+y^2)^(1/2) < δ.