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.
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.......