Quiz 7
1. Let x = r cos θ and y = r sin θ, where θ ∈ [0, 2π) and r ≥ 0. Note that (x, y) → 0 is equivalent
to r → 0+ , so
2x2 y
(x,y)→(0,0) x2 + y 2
lim
lim 2r cos2 θ sin θ.
=
r→0+
Notice that for any path as (x, y) goes the origin (i.e., for any values of θ),
−2r ≤ 2r cos2 θ sin θ ≤ 2r.
By the squeeze theorem,
lim
2x2 y
= 0.
+ y2
(x,y)→(0,0) x2
2. As you approach the origin along the x-axis (i.e., when y = 0),
x2 − y 2
=
lim
1 = 1,
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0)
lim
and when you approach the origin along the y-axis (i.e., when x = 0),
x2 − y 2
=
lim
−1 = −1.
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0)
lim
Since approaching the origin from different directions gives different values, the limit DNE.
3. (a) Make friends with the product rule and the fact that
i
∂ h x/y
xe
+ cos(y 2 )
∂x
=
∂
∂x
cos(y 2 ) = 0.
(1 + x/y) ex/y .
(b) Fun with the chain rule:
i
∂ h x/y
xe
+ cos(y 2 )
∂y
= −
x2 x/y
e
− 2y sin(y 2 ).
y2