1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solve this multivariable limit using epsilon-delta methods

  1. Sep 25, 2012 #1
    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?
     
    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 25, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Expand the difference of cubes.

    x3 - y3 = (x - y)(x2 + xy + y2)

    Change to polar coordinates.
     
  4. Sep 25, 2012 #3
    How could I go about it without the use of polar coordinates? I would like to see the different ways to solve this.
     
  5. Sep 25, 2012 #4

    Zondrina

    User Avatar
    Homework Helper

    Well, you know that |x| < δ and |y| < δ from the definition... use this to your advantage.
     
  6. Sep 25, 2012 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I would think of the squeeze theorem.
     
  7. Sep 25, 2012 #6
    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.
     
  8. Sep 25, 2012 #7

    Zondrina

    User Avatar
    Homework Helper

    Think about what you're saying here...

    Remember also that |x| ≤ (x^2+y^2)^(1/2) in this scenario...
     
  9. Sep 25, 2012 #8
    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?
     
  10. Sep 25, 2012 #9

    Zondrina

    User Avatar
    Homework Helper

    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.
     
  11. Sep 25, 2012 #10
    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.
     
  12. Sep 25, 2012 #11

    Zondrina

    User Avatar
    Homework Helper

    [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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solve this multivariable limit using epsilon-delta methods
Loading...