Continuity and Limits of Functions Exercise Answers
1. Let
f
be given by
f
(
x
) =
√
4

x
for
x
≤
4 and let
g
be given by
g
(
x
) =
x
2
for all
x
∈
R
.
(a) dom(
f
+
g
) = dom(
fg
) = (
∞
,
4], dom(
f
◦
g
) = [

2
,
2] and dom(
g
◦
f
) =
(
∞
,
4]
(b) (
f
◦
g
)(0) = 2, (
g
◦
f
)(0) = 4, (
f
◦
g
)(1) =
√
3,(
g
◦
f
)(1) = 3, (
f
◦
g
)(2) = 0 and
(
g
◦
f
)(2) = 2.
(c) No!
(d) (
f
◦
g
)(3) is not, but (
g
◦
f
)(3) is.
2. Let
f
be given by
f
(
x
) = 4 for
x
≥
0,
f
(
x
) = 0 for
x <
0 and let
g
be given by
g
(
x
) =
x
2
for all
x
∈
R
.
(a)
f
+
g
:
R
→
R
is given by (
f
+
g
)(
x
) = 4 +
x
2
for all
x
≥
0 and (
f
+
g
)(
x
) =
x
2
for all
x <
0.
fg
:
R
→
R
is given by (
fg
)(
x
) = 4
x
2
for all
x
≥
0 and (
fg
)(
x
) = 0 for all
x <
0.
f
◦
g
:
R
→
R
is given by (
f
◦
g
)(
x
) = 4 for all
x
∈
R
.
g
◦
f
:
R
→
R
is given by (
g
◦
f
)(
x
) = 16 for all
x
≥
0 and (
g
◦
f
)(
x
) = 0 for
all
x <
0.
(b) The functions
g
,
fg
and
f
◦
g
, are continuous, while
f
,
f
+
g
and
g
◦
f
are
discontinuous because they are discontinuous at 0. Note that although
f
is
discontinous, the function
fg
is continuous – so the continuity of
fg
does
not
imply both
f
and
g
are continuous.
3. The functions given by sin
x
, cos
x
,
e
x
, 2
x
, ln
x
for
x >
0, and
x
p
for
x >
0 (
p
∈
R
)
are continuous on their domains. Use these facts and theorems in the notes to prove
that the functions given as below are also continuous.
1
(a) ln(1 + cos
4
x
).
Solution.
First 1 =
x
0
is continuous by one of the above facts. Next, cos
x
is
continuous from above, and cos
4
x
= (cos
x
)
4
is a composition of two functions
(given by cos
x
and
y
4
) which are continuous on their domains. Thus, since the
composition of two continuous functions is continuous, cos
4
x
is continuous.
Now, since a function formed by pointwise addition is continuous, it follows
that (1 + cos
4
x
) is a continuous function. Finally, since cos
4
x
≥
0 for all
x
∈
R
, we have 1 + cos
4
x
≥
1
>
0 for all
x
∈
R
. Thus since ln
x
is continuous
for
x >
0, the composition on the right of ln with (1+cos
4
x
) is continuous.
1
We have used cos
n
x
to denote (cos
x
)
n
for any
n
∈
N
and similarly for sin
n
x
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(b) [sin
2
x
+ cos
6
x
]
π
(c) 2
x
2
(d) 8
x
(e) tan
x
for
x
6
= odd multiple of
π/
2
(f)
x
2
sin(1
/x
) for
x
6
= 0
(g)
x
2
sin(1
/x
) for
x
6
= 0
(h) (1
/x
) sin(1
/x
2
) for
x
6
= 0
4. Prove that the function
√
x
is continuous on its domain [0
,
∞
).
Hint
: use the se
quential definition of continuity and the fact that if (
s
n
) is a sequence of nonnegative
real numbers and
s
= lim
s
n
, then lim
√
s
n
=
√
s
.
5.
(a) Prove that if
m
∈
N
, then the function
f
(
x
) =
x
m
is continuous on
R
.
Hint
:
You can construct an
ε

δ
proof using the identity
x
m

y
m
= (
x

y
)(
x
m

1
+
x
m

2
y
+
· · ·
+
xy
m

2
+
y
m

1
)
.
Or you can prove the result using induction on
m
. First prove that
g
(
x
) =
x
is continuous on
R
. Then use an inductive hypothesis that
f
(
x
) =
x
m
is
continuous on
R
and the theorem about the continuity of (
fg
).
Solution.
Let
ε >
0 and let
x
0
be an arbitrary real number. We need to find a
δ >
0 such that if

x

x
0

< δ
then

x
m

x
m
0

< ε
. Now, by the hint,

x
m

x
m
0

=

x

y

x
m

1
+
x
m

2
x
0
+
· · ·
+
xx
m

2
0
+
x
m

1
0

.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Continuity, Limits, Continuous function, Xn, lim sgn

Click to edit the document details