Homework due 7th February
Solutions to compulsory questions
Problem (
§
2.2 Exercise 8b).
Compute the limit
lim
(
x,y
)
→
(0
,
0)
sin(
xy
)
y
if it exists.
Solution.
Using the substitution
u
=
xy
we have
lim
(
x,y
)
→
(0
,
0)
sin(
xy
)
(
xy
)
= lim
u
→
0
sin
u
u
= 1
.
Therefore, by the properties of limits,
lim
(
x,y
)
→
(0
,
0)
sin(
xy
)
y
=
lim
(
x,y
)
→
(0
,
0)
±
x
×
sin(
xy
)
(
xy
)
²
=
±
lim
(
x,y
)
→
(0
,
0)
x
²±
lim
(
x,y
)
→
(0
,
0)
sin(
xy
)
(
xy
)
²
= 0
.
Alternatively, by expanding sin(
xy
) into a Taylor series we have
lim
(
x,y
)
→
(0
,
0)
sin(
xy
)
y
=
lim
(
x,y
)
→
(0
,
0)
(
xy
)

1
3!
(
xy
)
3
+
···
y
=
lim
(
x,y
)
→
(0
,
0)
x
±
1

(
xy
)
2
3!
+
···
²
= 0
.
Problem (
§
2.2 Exercise 10c).
Compute the limit
lim
(
x,y
)
→
(0
,
0)
(
x

y
)
2
x
2
+
y
2
if it exists.
Solution.
Along the line
y
=
x
we have
f
(
x, x
) =
(
x

x
)
2
x
2
+
x
2
= 0
x
→
0
→
0
,
and along the line
y
=

x
we have
f
(
x,

x
) =
(
x
+
x
)
2
x
2
+
x
2
= 2
x
→
0
→
2
.
Therefore the limit does not exist.
Problem (
§
2.2 Exercise 15).
If
f
:
R
n
→
R
and
g
:
R
n
→
R
are continuous, show that the
function
f
2
+
g
:
R
n
→
R
x
7→
[
f
(
x
)]
2
+
g
(
x
)
is continuous.
Solution.
If
a
∈
R
n
then
lim
x
→
a
(
f
2
+
g
)(
x
) = lim
x
→
a
(
[
f
(
x
)]
2
+
g
(
x
)
)
(deﬁnition of
f
2
+
g
)
=
³
lim
x
→
a
f
(
x
)
´
2
+
³
lim
x
→
a
g
(
x
)
´
(using limit properties)
= [
f
(
a
)]
2
+
g
(
a
)
(since
f, g
continuous)
= (
f
2
+
g
)(
a
)
(deﬁnition of
f
2
+
g
)
.
Therefore
f
2
+
g
is continuous at each
a
∈
R
n
, and so
f
2
+
g
is continuous on
R
n
.
1