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Limit of polar coordinates

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known dataI need to evaluate this limit by converting to polar coordinates:

    lim (x,y) -> (0,0) of (x^2 + xy + y^2) / x^2 + y^2



    2. Relevant equationsx = rcos(theta), y = rsin(theta)



    3. The attempt at a solutionSo switching to polar I get:

    [(rcos(theta))^2 + rcos(theta)rsin(theta) + (rsin(theta))^2] / (rcos(theta))^2 + (rsin(theta))^2

    By pulling out the r^2 from the the top of the equation and the bottom of the equation, they can cancel. Then the denominator is cos(theta)^2 + sin(theta)^2 which equals 1.

    So we get the limit of cos(theta)^2 + cos(theta)sin(theta) + sin(theta)^2 but I dont know what to do from here because this is the limit as r goes to 0 and there is no r?

    I'm kinda stuck here...what can I do? We didn't really get taught this so I could be missing something simple.

    Thanks!
     
  2. jcsd
  3. Sep 18, 2008 #2

    Dick

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    You can simplify a little more down to 1+cos(theta)sin(theta). There is always the possible answer that the limit doesn't exist, right? What do you say in this case and why?
     
  4. Sep 18, 2008 #3
    Right I figured the limit did not exist. Does it have to do wiht the cos(theta)sin(theta)? So As r goes to 0, the function is just 1cos(theta)sin(theta) for whatever value of theta which will oscillate. Is that the correct way of thinking about it?
     
  5. Sep 18, 2008 #4

    Dick

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    Look at it this way. For the limit to exist the limit has to be independent of the way (x,y) approaches (0,0). If you set y=0, and let x->0, what's the limit. (This is the theta=0 case, right? Check it in your polar formula.) Now set y=x and let x->0. (This is the theta=pi/4 case. Try putting that into your polar formula as well.). So right, the limit does not exist. Because it depends on theta.
     
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