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Solve this multivariable limit using epsilon-delta methods

  • Thread starter MeMoses
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  • #1
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Homework Statement



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

Homework Equations





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?
 
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  • #2
SammyS
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Homework Statement



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

Homework Equations



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.
 
  • #3
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How could I go about it without the use of polar coordinates? I would like to see the different ways to solve this.
 
  • #4
Zondrina
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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.
 
  • #5
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I would think of the squeeze theorem.
 
  • #6
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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.
 
  • #7
Zondrina
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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...
 
  • #8
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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?
 
  • #9
Zondrina
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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.
 
  • #10
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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.
 
  • #11
Zondrina
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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.
 

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