Worksheet-11
10/7/2014
Please work in groups and use the given time to think and try to do the problems. If you can not
solve a question please ask me for a hint or just pass that question.
1. Use limit laws to evaluate the following limits.
(a)
lim
(x2 + y)
(x,y)→(1,2)
Solution:
(x2 + y) =
lim
(x,y)→(1,2)
(b)
lim
x2 +
(x,y)→(1,2)
y = 12 + 2 = 3.
lim
(x,y)→(1,2)
2x2
(x,y)→(−2,1) 4x + y
lim
2x2
Solution:
lim
=
(x,y)→(−2,1) 4x + y
2x2
lim
(x,y)→(−2,1)
lim
4x + y
=
8
−8
=
−8 + 1
7
(x,y)→(−2,1)
2. Let f (x, y) = xy/(x2 + y 2 ). Show that f (x, y) approaches zero along the x-axes and the
y-axes. Then prove that
lim f (x, y) does not exist by showing that the limit along the
(x,y)→(0,0)
line y = x is nonzero.
Solution: Along the x-axis we have:
lim
(x,y)→(0,0)
xy
0
=
lim
=
lim 0 = 0.
x2 + y 2 (x,y)→(0,0) x2 (x,y)→(0,0)
Along the y-axis we have:
lim
(x,y)→(0,0) x2
0
xy
=
lim
=
lim 0 = 0.
2
+y
(x,y)→(0,0) y 2
(x,y)→(0,0)
But Along the line y = x we have:
lim
(x,y)→(0,0) x2
x2
1
1
xy
=
lim
=
lim
= ,
2
2
+y
2
(x,y)→(0,0) 2x
(x,y)→(0,0) 2
so the limit DNE.
3. Evaluate the following limit or determine why it doesn’t exist.
(a)
x2 − y 2
(x,y)→(0,0) x2 + y 2
Solution: Along the x-axis we have:
lim
x2 − y 2
x2
=
lim
=
lim 1 = 1.
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0) x2
(x,y)→(0,0)
lim
Along the y-axis we have:
x2 − y 2
−y 2
=
lim
=
lim −1 = −1.
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0) y 2
(x,y)→(0,0)
lim
So the limit DNE.
1
(b)
xy 4
(x,y)→(0,0) x2 + y 2
Solution: We notice that the limit is 0 along the x-axis, y-axis and y = x. So we look
for an algebraic approach. We convert to polar coordinates and get:
lim
xy 4
r cos θ(r sin θ)4
r5 cos θ sin4 θ
=
lim
=
lim
= lim r3 cos θ sin4 θ = 0,
r→0
r→0
r→0
r2
r2
(x,y)→(0,0) x2 + y 2
lim
since sin4 θ and cos θ are both bounded functions.
xy
p
(c)
lim
(x,y)→(0,0)
x2 + y 2
Solution: We notice that the limit is 0 along the x-axis, y-axis and y = x. So we look
for an algebraic approach. We convert to polar coordinates and get:
r2 cos θ sin θ
xy
r cos θr sin θ
p
= lim
= lim r cos θ sin θ = 0,
= lim
r→0
r→0
r
r
(x,y)→(0,0)
x2 + y 2 r→0
lim
since sin θ and cos θ are both bounded functions.
1
(d)
lim tan(x2 + y 2 ) arctan( 2
)
x + y 2 (x,y)→(0,0)
Solution: Since
−π
2
≤ arctan
1
x2 +y 2
≤
π
2,
we will use the squeeze theorem on this
1
tan(x2 + y 2 ) ≤ tan(x2 + y 2 ) arctan x2 +y
≤ π2 tan(x2 + y 2 ).
2
limit. We notice that −π
2
Also:
−π
−π
−π
lim
tan(x2 + y 2 ) =
tan(02 + 02 ) =
tan(0) = 0 and
2
2
(x,y)→(0,0) 2
π
π
π
lim
tan(x2 + y 2 ) = tan(02 + 02 ) = tan(0) = 0, so by the squeeze theorem we
2
(x,y)→(0,0) 2
2
1
=0
have
lim tan(x2 + y 2 ) arctan
x2 + y 2
(x,y)→(0,0)
(e)
lim
(x,y)→(0,0)
x2 + y 2
p
x2 + y 2 + 1 − 1
Solution:
p
p
2 + y2)
2 + y2 + 1 + 1
(x
x
2
2
x +y +1+1
+
p
lim
·p
=
lim
2
2
x2 + y 2 + 1 − 1
(x,y)→(0,0)
x +y +1−1
x2 + y 2 + 1 + 1 (x,y)→(0,0)
p
x2 + y 2 + 1 + 1
(x2 + y 2 )
=
lim
x2 + y 2
(x,y)→(0,0)
p
=
lim
x2 + y 2 + 1 + 1
(x,y)→(0,0)
p
= 02 + 0 2 + 1 + 1
x2
y2
=2
4. Compute the following partial derivatives:
(a)
∂
∂y (sin (x) sin (y))
∂
Solution: ∂y
(sin (x) sin (y))
= sin(x) cos(y)
2
(b)
∂ xy
∂y x−y
Solution:
(x − y)(x) − (xy)(−1)
x2 − xy + xy
x2
∂ xy
=
=
=
2
2
∂y x − y
(x − y)
(x − y)
(x − y)2
(c)
∂ 2 xy
∂x∂y x−y
Solution:
∂2
xy
∂ xy
∂
x2
∂
=
=
∂x∂y x − y
∂x ∂y x − y
∂x (x − y)2
2
2
(x − y) (2x) − x (2(x − y))(1)
(x2 − 2xy + y 2 )2x − (2x3 − 2x2 y)
=
=
(x − y)4
(x − y)4
2x3 − 4x2 y + 2xy 2 + 2xy 2 − 2x3 + 2x2 y
4xy 2 − 2x2 y
=
=
(x − y)4
(x − y)4
(d)
√
∂
u2 +v 2 .
e
∂u
Solution:
√
∂ √u2 +v2
2
2
e
= e u +v
∂u
√
2
2
1 2
ue u +v
2 −1/2
(u + v )
(2u) = √
2
u2 + v 2
5. Verify Clairaut’s Theorem for the function z = x4 cos xy. In other words, show that zxy = zyx .
Solution: First we find zxy .
zxy = (zx )y = x4 − sin(xy)y + 4x3 cos(xy) y
= 4x3 cos(xy) − x4 y sin(xy) y
= 4x3 − sin(xy)x − x4 sin(xy) − x4 y cos(xy)x
= −4x4 sin(xy) − x4 sin(xy) − x5 y cos(xy)
= −5x4 sin(xy) − x5 y cos(xy)
Next we find zyx .
zyx = (zy )x = (x4 − sin(xy)x)x
= (−x5 sin(xy))x
= −5x4 sin(xy) − x5 cos(xy)y
= −5x4 sin(xy) − x5 y cos(xy)
So we see that in both cases we get the same final result.
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