Limit of multivariable function

1. Oct 12, 2008

alok24590

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
!(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.......

2. Oct 12, 2008

HallsofIvy

Staff Emeritus
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.