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Limit of multivariable function

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    lim(x,y)->(4,pi)(x^2sin(x/y)), i tried it by polar as well as epsilon delta method, but by polar coordinate method it is coming that " limit doesn't exits" but by epsilon delta method "it exists & is equal to (epsilon+(16/(2)^(1/2)) )^(1/2)-4.

    2. Relevant equations

    3. The attempt at a solution
    replacing x by rcos(alpha)& y by rsin(alpha) i get lim(r)->((16+pi^2)^(1/2))(rcos(alpha))^(2).sin(cot(alpha)) & since it depends upon alpha hence limit of it doesn't exists. but by epsilon delya method
    let limit is equal to 16/(2)^(1/2)
    the limit can be replaced as lim(z,k)->(0,0)((z+4)^(2)sin((z+4)/(k+pi)))
    take any epsilon>0
    to show that there exists a delta>0 such that !(z,k)!<delta
    it empiles !f(z,k)-(16/(2)^(1/2))!<epsilon
    now replacing z by rcos(alpha)& k by rsin(alpha)we get
    that is indeed <(r+4)^(2)-(16/(2)^(1/2))
    replacing r by delta we get we get the answer that i told earlier.......
  2. jcsd
  3. Oct 12, 2008 #2


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    Staff Emeritus
    Science Advisor

    The limit is at (4, pi) so r alone does NOT measure the distance of (x,y) to (4, pi). That method only works if the limit is at (0,0) so that r does measure that distance, regardless of theta. What you did in your second method was to "translate" the coordinate system so that (0,0) in (z, k) is (4, pi) in (x,y). Now that method works.
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