Graphics Reference
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(iii) Of the
m
(
m
1) or less numbers,
a
p
/
q
, noneof which
are zero, there is a least.
(iv) If
x
is irrational, then
f
(
x
)
f
(
a
)
0.
If
x
is rational and
x
a
, then
x
p
/
q
with
q
m
.
So
f
(
x
)
f
(
a
)
f
(
x
)
1/
q
1/
m
.
73 For example
a
1
(
)
. Eventually
a
1.
If
a
1, then (
a
)
1, but (
f
(
a
))
1.
74 0, 1,
1.
75 If for
every
sequence (
a
)
a
,(
f
(
a
))
f
(
a
), then for every sequence
(
a
)
a
with
a
a
,(
f
(
a
))
f
(
a
).
76 Answer in summary.
77 1, 0,
1.
78 See answer to qn 75.
79 If
f
were continuous at
a
then both limits would equal
f
(
a
) from qns 75
and 78.
80
1, 1.
81
(i)
0.005.
0.0005.
(iii) Yes.
(ii)
1.895,
0.551,
0.173.
82 Let
f
(
x
)
1 when
x
0, and let
f
(0)
2.
83 Let (
a
)
a
where
a
a
. Given
0, eventually
a
a
a
,so
l
.
So the neighbourhood definition of limit implies the sequence definition
of limit.
f
(
a
)
l
and this proves that (
f
(
a
))
84 From
a
a
a
1/
n
,(
a
)
a
by qn 3.54(viii), the squeeze rule. But
)) does not tend to
l
, which contradicts
our hypothesis. So the sequence definition of limit implies the
neighbourhood definition of limit.
f
(
a
)
l
implies that (
f
(
a
85 Answer in summary.
87 Given
0, thelimit from blow says that, for some
,
a
,
and thelimit from abovesays that, for some
x
a
f
(
x
)
l
,
a
x
a
f
(
x
)
l
.
So if 0
x
a
min(
,
), it follows that
f
(
x
)
l
.
88
(i)
a
x
a
0
x
a
f
(
x
)
l
.
.
(ii) The two conditions for one-sided limits are obviously satisfied.
a
x
a
0
x
a
f
(
x
)
l
