Graphics Reference
In-Depth Information
28 Let
a
x
, then
1
x
1
a
1
a
a
...
a
f
(
x
)
.
1
x
1
a
1
a
a
...
a
As
x
1,
a
1, by thecontinuity of
x
x
.
Now by qns 6.29, 6.54 and 6.86 lim
f
(
x
)
n
/
m
.
29 Therangemust bean intrval, by qn 21.
30 The only possible ranges are a singleton and a closed interval.
b
.
(iii) Yes, by qn 4.46, every bounded sequence has a convergent
subsequence.
(iv) By qn 3.78, the closed interval property.
(v) Its limit is
f
(
c
).
(ii)
a
x
31
(i) None whatsoever.
(vi) Take
1.
(vii) Choose
n
f
(
c
)
1, then, from the definition of
x
,
f
(
x
f
(
c
)
1, and this contradicts (vi), so our original
hypothesis was wrong.
)
n
32 As in qn 31, but starting from a sequence (
x
) in thedomain of the
function such that
f
(
x
)
n
.
33
(i) Denominator not zero on
Q
, so thefunction is continuous by
qn 6.51.
(ii) Because
need
not contain a subsequence which converges to a point of
Q
is not complete, a bounded sequence in
Q
Q
.
34
(i) From qn 21.
(ii) From qn 31.
(iii) Every non-empty set of real numbers which is bounded above
has a least upper bound: qns 4.80 and 4.81.
(iv) Question 4.64.
(v) Since
a
x
) is bounded, and so must contain a
convergent subsequence by qn 4.46, every bounded sequence
has a convergent subsequence.
(vi) By qn 3.78, the closed interval property.
(vii) Because
f
is continuous.
(viii) Since
b
,(
x
f
(
x
)
M
1/
n
, the limit of the subsequence (
f
(
x
)) is
M
,so
M
f
(
c
).
(ix) If inf
V
m
, construct a sequence (
y
) such that
m
f
(
y
)
m
1/
n
. Argueas in (iv)
—
(viii).
35 sin
x
1 only when
x
(
n
)
which is irrational. But the
domain of thefunction is densein
R so it attains values arbitrarily
closeto
1.
Since
Q
is not complete, the argument of qn 34(v) fails on
Q
.
