Graphics Reference
In-Depth Information
Answers
1
(i) The definition gives the basis for the induction.
Then
x
.
(ii) The definition gives the basis for the induction.
x
x
x
x
(
x
)
x
(
x
)
.
(iii) (
xy
)
(
xy
)
(
xy
)
x
y
xy
x
xy
y
x
y
.
x
x
x
x
x
x
x
2
m
n
0
n
m
1
x
x
x
1/
x
1/
x
.
3 See qn 2.20 for strict increase, qn 6.28 for continuity. 1
x
x
shows
unboundedness above.
4 See qn 7.27, continuity of
n
th root function.
5
(i) Use1(iii) and thefact that
x
x
is a bijection.
(ii) (
x
)
(
x
)
x
(
x
)
x
(
x
)
(
x
)
x
(
x
x
)
(
x
)
.
6 Let
r
p
/
m
and
s
q
/
n
, where
m
,
n
,
p
,
q
Z
.
(i)
x
x
(
x
)
(
x
)
(
x
)
x
x
x
x
(ii)
x
x
(
x
)
(
x
)
, using qn 5(ii) and 5(iii).
7
r
s
0
s
r
1
x
x
x
1/
x
1/
x
. Use3.57 and 2.23,
reciprocal inequalities.
8 The argument of qn 6 applies. For negative
r
and
s
, work with
reciprocals.
9 The argument of qn 7 applies.
10
A
(
x
)
1/
A
(
x
). For some
N
,
m
N
A
(
1/
m
)
1
, and
because
A
is strictly increasing,
1/(
N
1)
x
1/(
N
1)
A
(
x
)
1
. Thus
A
(
x
)
1
A
(0) as
x
0.
11
x
q
x
q
A
(
x
q
)
A
(
x
q
)
0
1
1
0
A
(
x
)
A
(
q
)
A
(
x
)
A
(
q
).
0
12 By qns 6.23, continuity of sums of functions, and 6.54, continuity of
quotients of functions.
13 Let
r
p
/
m
and
s
q
/
n
, where
m
,
n
,
p
,
q
Z
.
r
s
pn
qm
D
(
pn
)
D
(
qm
). Now put
b
a
, and wegt
D
(
r
)
D
(
s
) but with
b
for
a
in thedefinition of
D
.
14 If
m
N
D
(1/
m
)
L
(
a
)
, then because
D
is strictly increasing on
Q
,0
x
1/(
N
1)
D
(
x
)
L
(
a
)
.
15 Apply qn 6.93, the algebra of limits, from above.
