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Using polar coordinates to determine the limit

  1. Jun 15, 2014 #1
    Lim (x, y)->(0,0)(X^3+y^3)/(x^2+y^2)

    The answer is -1, but I can't get it there. Here is what I did.
    ((Rcosx)^3 +(rsinx)^3)/((rcosx)^2+(rsinx)^2)
    Then by factoring out a r squared from top and bottom I'm left with a denominator of (sin^2(x ) + cos^2 (x)) which simplifies to 1. And a numerator of:
    Rcos^3(x)+rsin^3 (x)

    This is about as far as I've gotten and I can't figure out how to get it to a -1.
     
  2. jcsd
  3. Jun 15, 2014 #2

    Zondrina

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    Homework Helper

    The answer is certainly not -1. The answer is 0 as ##r → 0##.

    Conversely, if ##\epsilon > 0## is given, then choose ##\delta = \frac{\epsilon}{2}##.
     
  4. Jun 15, 2014 #3

    ehild

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    Gold Member

    The limit is not -1. If it were, the function should get closer and closer to it, when (x,y) gets closer and closer to (0,0). If you choose x=y, f(x,x)=x →0. If you choose x=rcosθ, y=rsinθ, f=r(cos3θ+sin3θ) which goes to 0 if r→0, for any value of θ.

    ehild
     
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