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Limit of function of two variables

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all real a and b such that the limit [tex]\lim_{(x,y) \to (0,0)} \frac{\left\vert x \right\vert ^a \left\vert y \right\vert ^b}{x^2+y^2}[/tex] exists and is finite. (Hint: it is not enough to find the limit or prove it exists in the region found. You must also prove that the limit is infinite or nonexistent outside the region, and also justify your decision about which boundary points to include.)


    2. Relevant equations



    3. The attempt at a solution

    Since this is a limit in 2D, I tried the common paths first.

    Substituting x = 0 gives 0 on the numerator (if a =/= 0), and y^2 on the denominator. I'm not sure how I could evaluate this limit further.

    Similarly for y = 0

    Substituting y = kx gives [tex]\lim_{x \to 0} \left\vert x \right\vert ^{a+b-2} \frac{\left\vert k\right\vert ^b}{k^2+1}[/tex] thus a + b - 2 > 0 to ensure the existence of the limit

    Substituting y = mx^2 gives [tex]\lim_{x \to 0} \left\vert x\right\vert ^{a+2b-2} \frac{\left\vert m \right\vert ^b}{1+m^2x^2}[/tex] thus a + 2b - 2 > 0

    then use the polar coordinates [tex]x=rcos\theta , y=rsin\theta[/tex] the limit becomes [tex]\lim_{r \to 0} r^{a+b-2} \left\vert cos\theta \right\vert ^a \left\vert sin\theta \right\vert ^b[/tex]. hence a + b - 2 > 0 as above

    I can't think of any other approaches to evaluate this limit though, and I have no idea how to show limit is nonexistent outside (a+b>2 and a+2b>2) and justify the boundary points chosen. It would be great if anyone could share their thoughts on this. Thanks in advance.
     
  2. jcsd
  3. May 17, 2010 #2

    Cyosis

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    Outside the region you have found that the exponent of r is negative. Investigate the different cases where the negative exponent is even and when it is odd.
     
  4. May 17, 2010 #3
    It suffices to show the exponent of base r is negative when a + b < 2? Do I need to examine other paths as approaches to evaluate the limit? Since the polar coordinate substitution seems to encompass more general cases, would I still need the a + 2b > 0 from the parabola path though?

    Also could you elaborate on the second part - about even and odd negative exponents. I can't really see the difference - wouldn't they all make the limit of r^a (with a < 0) as r approaches 0 become nonexistent? Thanks.
     
  5. May 20, 2010 #4

    Gib Z

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    This is an ASSIGNMENT question from the MATH1901 course at the University of Sydney, and therefore a violation of PF guidelines to not at least inform us. Please refrain from helping this user, this assignment counts towards the final mark.
     
  6. May 20, 2010 #5
    Please delete this post if this constitutes a violation of the forum rules. I did not realise that. Sorry.
     
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