Graphics Reference
In-Depth Information
x
a
or
x
a
x
a
89 0
.
is trivial when
x
a
.
First proposition is equivalent to the limit (qn 88).
Second proposition implies continuity (qn 67).
90
k
f
(
x
)
f
(
a
)
2. Calculatelimits from aboveand blow
x
1 and useqns 79
and 89.
91
f
(
a
) from qn 78,
f
(
b
) from qn 75 and
f
(
c
) from qn 86.
92 Continuous at
c
from qn 89.
If (
a
) is any sequence in [
a
,
c
] which tends to
a
, it has either a finite or
an infinite number of terms different from
a
. If it only has a finite
number of terms different from
a
,(
f
(
a
)) is eventually constant with
f
(
a
). If it has an infinitenumbr of
terms different from
a
, and (
a
terms equal to
f
(
a
), so (
f
(
a
))
) is the subsequence of such terms, then
(
f
(
a
))
f
(
a
) for the resulting subsequence by the one sided limit. In
either case, (
f
(
a
))
f
(
a
).
Continuity at
b
follows similarly.
93 First result from qn 3.54(vii), the absolute value rule, second from
qn 3.54(iii), thesum rul, third from qn 3.54(vi), theproduct rul,
and the fourth from qns 3.65 and 3.66, the reciprocal rule.
94 The development is like that for continuous functions in qns 21, 22,
23, 25, 26, 28, 29, and then uses qn 93 parts three and four for
rational functions. Composite functions must not be used; see qn
98.
95 Uselimits from aboveand blow
x
2. Take
k
4.
96
Since
f
(
b
)
g
(
b
), lim
F
(
x
)
f
(
b
)
g
(
b
)
lim
F
(
x
), so
F
is
continuous at
b
, by qns 87 and 89. For
x
b
, any sequence tending
to
x
[
a
,
c
] is eventually in [
a
,
b
]or[
b
,
c
] and so
F
is continuous at
b
from thecontinuity of
f
or
g
.
97
x
a
(
x
a
)(
x
ax
...
a
x
a
).
So lim
f
(
x
)
na
k
.
98
f
(
x
)
l
g
(
x
)
l
h
(
x
)
l
, for
x
a
.
If 0
g
(
x
)
l
, then
g
(
x
)
l
h
(
x
)
l
.
If
g
(
x
)
l
0, then
g
(
x
)
l
f
(
x
)
l
.
Given
0, there is a
such that
0
x
a
f
(
x
)
l
, and there is a
such that
0
x
a
h
(
x
)
l
.
So, if 0
x
a
min(
),
g
(
x
)
l
max(
f
(
x
)
l
,
h
(
x
)
l
)
.
,
Thersult follows from qn 87.
