x2 y2
(from several perspectives). You can see from the generated graph that
x 2+ y 4
it goes through the origin, so the point is well-defined there.
Graph of surface z=
So, we already had
cosθ≠0 .
lim
( x , y) →(0,0 )
r 4 (cos 2 θsin 2 θ)
r 2 (cos2 θsin 2 θ)
x2 y2
=lim
=lim
=0 as long as
x 2 + y 4 r → 0 r 2 (cos 2 θ+r 2 sin 2 θ) r → 0 cos2 θ+r 2 sin 2 θ
And, if cos θ=0 , that's equivalent to θ=
as θ →
( 2n+1)π
, n∈ ℤ . So, we can figure out what happens
2
(2n+1)π
.
2
r 2 (cos 2 θ sin 2 θ)
is the 0/0 case, so we can use L'Hopital's Rule to get
lim
2
2
2
(2n +1) π cos θ+r sin θ
θ→
2
lim
=
θ→
(2n +1) π
2
−2r 2 sin 2 θ+2r 2 cos 2 θ
−2+4r 2 sin 2 θ
(after simplifying) =
−2r 2
r2
=
and as r goes to
−2+4r 2 1−2r 2
zero here, we still get 0. This completes the proof that the limit exists and goes to 0.