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In-Depth Information
L
(
D
(
q
))
(
a
). The result comes from qn 3.54(viii), the squeeze
rule.
(v) From (i) and qn 6.89, continuity by limits.
24 (
A
(
x
)
A
(
x
))/(
x
x
)
A
(
x
)·
D
(
x
x
)
A
(
x
)·
L
(
a
)as
x
x
.
25 Both (
D
(1/
n
)) and (
D
(
1/
n
))
L
(
a
)as
n
.
26 0
L
(
a
)
a
1, so
L
(
a
)
0as
a
1
.
1
n
For
a
2,
L
(
a
)
as
a
, by qn 5.30, theharmonic
series.
L
is strictly increasing because
x
0
1/
x
0.
L
is continuous by qn 10.50, thecontinuity of integrals.
L
(
x
)
1/
x
by
the Fundamental Theorem of Calculus.
27 Since
is continuous and takes values from small positive to large
positiveon (1,
L
takes the value 1 by the Intermediate Value
Theorem for some value of
a
in this domain. On (1, 2), 1/
x
),
L
1, so
L
(2)
1.
28 Follows from
L
(e)
1 and qn 24. Taylor series from qn 9.38.
29
f
(
x
)
L
(
E
(
x
))·
E
(
x
)
(1/
E
(
x
))·
E
(
x
)
1. Thus
f
(
x
)
x
c
. But
f
(0)
0, so
c
0.
30 Since
E
L
,
E
(
L
(
x
))
x
, and
L
(
x
)
log
x
.
31
(i) First equality by definition of log. Second equality from second
law of indices.
(ii) Thelogarithm of theequality in (i). Useqn 29.
(iii)
E
(log
b
)
b
.
E
(log
b
· log
a
)
A
(log
b
) by (i),
b
by definition.
32 The function is well defined provided
x
1/
a
.
a
1
ax
a
. Then consider
x
1/
n
.
Limit
lim
34
f
(
x
)
E
(
a
log
x
).
f
(
x
)
E
(
a
log
x
)· (
a
/
x
)
f
(
x
)(
a
/
x
).
35 Let
f
(
x
)
x
e
.
f
(
n
1)/
f
(
n
)
(1
1/
n
)
/e
1/eas
n
.Byqn
3.71, d'Alembert, (
f
(
n
))
0as
n
.
f
(
x
)
f
(
x
)(
a
/
x
1)
0 when
x
a
.So
f
is strictly decreasing when
x
a
. Since the sequence
0, thefunction
0as
x
.So
1/
f
(
x
)
.
1
a
log
x
Let
g
(
x
)
(log
x
)/
x
, then
g
(
x
)
0 when
x
e
.
x
Let
h
(
n
)
g
(e
), then
h
(
n
1)/
h
(
n
)
(1
1/
n
)/e
1/e
as
n
. So,
