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Find the limit

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the limit:

    the limit of (x2y)/(x4 + y2) as (x,y) approaches (0,0)


    2. Relevant equations



    3. The attempt at a solution

    I took the limit of the numerator and denominator separately.

    The numerator equals to 0 as well as the denominator.

    So, I get the indeterminant form 0/0.

    Where do I go from here?
     
  2. jcsd
  3. Apr 9, 2014 #2
    Try approaching the problem using polar coordinates:

    [itex]x=r\cos{\theta}[/itex]
    [itex]y=r\sin{\theta}[/itex]
     
  4. Apr 9, 2014 #3

    After solving I get cos2(theta) / sin(theta)
     
  5. Apr 9, 2014 #4

    Zondrina

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    You could choose ##\delta = min\{1,2 \epsilon\}##
     
  6. Apr 9, 2014 #5
    Uh according to what I have, I'll do the numerator and let you do the denominator:

    [itex]
    x^2 y = r^3 \cos{\theta}^2\sin{\theta}
    [/itex]

    Also remember [itex](x,y)\to 0[/itex] is the same as [itex]r\to 0[/itex]
     
  7. Apr 9, 2014 #6
    Can I choose y = 0 and find the limit. Then, y = x2 and find the limit. Will that show that it approaches different limits and Therefore, it would not exist?
     
  8. Apr 9, 2014 #7
    I get the limit to exist. Make sure you are making the substitutions correctly.
     
  9. Apr 9, 2014 #8
    denominator:

    x4 + y2 = r4cos4(theta) + sin2(theta)
     
  10. Apr 9, 2014 #9
    Ah see you are missing the r squared term in front of your sine term.
     
  11. Apr 9, 2014 #10

    SammyS

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    I get that the limit doesn't exist.
     
  12. Apr 9, 2014 #11
  13. Apr 9, 2014 #12

    SammyS

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    Let (x, y) → (0, 0) along either of the paths, y = x2 and y = -x2 .
     
  14. Apr 9, 2014 #13
    Touché, WolframAlpha lies, as my polar coordinate method is not sufficient.
     
  15. Apr 9, 2014 #14

    Ray Vickson

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    You shouldn't. You made a mistake.
     
  16. Apr 9, 2014 #15

    SammyS

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    I stumbled on that path quite by accident, although I did use WolframAlpha to help.

    I was somewhat surprised that WolframAlpha messed up evaluating this limit.
     
  17. Apr 10, 2014 #16
    Yeah I just used the polar coordinates method and assumed Wolfram would have given the correct answer.
     
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