Limit of multivariable function

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Homework Statement


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.



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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
!(rcos(alpha)+4)^(2).sin((rcos(alpha)+4)/(kcos(alpha)+pi))-(16/(2)^(1/2))!
that is indeed <(r+4)^(2)-(16/(2)^(1/2))
replacing r by delta we get we get the answer that i told earlier.......
 

Answers and Replies

  • #2
HallsofIvy
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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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