Graphics Reference
In-Depth Information
17
f
(
x
)/
g
(
x
)
f
(
a
)/
g
(
a
)
x
a
f
(
x
)·
g
(
a
)
f
(
a
) ·
g
(
x
)
g
(
x
) ·
g
(
a
) · (
x
a
)
g
(
a
)(
f
(
x
)
f
(
a
))
f
(
a
)(
g
(
x
)
g
(
a
))
g
(
x
) ·
g
(
a
) ·(
x
a
)
which gives the required result by qn 6.93, the algebra of limits, and the
continuity of 1/
g
at
a
.
g
(
x
)
g
(
a
)
18
(ii) lim
a
g
(
a
).
x
If, in every neighbourhood of
a
, there are points
x
a
such that
g
(
x
)
g
(
a
)
then
g
(
a
)
0: because,
G
(
a
)
0 for all
a
A
19 First limit
1, second limit
1, so two-sided limit does not exist.
20 If
a
1/(2
n
) then
f
(
a
)
0. If
b
1/((2
n
)
) then
f
(
b
)
1.
So (
f
(
a
)) have different limits and therefore
f
is not continuous
at 0 whatever the value of
f
(0). Comparewith qn 6.19.
)) and (
f
(
b
21
x
x
is continuous by qn 6.32.
x
x
is continuous, by qns 6.32 and 6.26.
So
f
is continuous, by qn 6.36. But
f
(0)
f
(
x
)
sin
1
0
x
, for
x
0,
x
and this has no limit as
x
0.
22
x
x
and
x
x
areboth continuous, by qn 6.29.
x
f
(
x
)
x
so
f
is continuous at
x
0, by qn 6.36.
f
(0)
x
f
(
x
)
0
x
sin
1
x
when
x
0as
x
0 and from qn 21 this
0.
f
(
a
h
)
f
(
a
)
h
23
If lim
f
(
a
),
f
(
a
h
)
f
(
a
)
then lim
f
(
a
). Add.
h
The result follows from qn 6.93, the algebra of limits.
The limit exists when
a
x
0 for
f
(
x
)
, but this function is not
differentiable at 0 by qn 19.
24 None of these functions is continuous when
x
0, using arguments like
that of qn 6.20.
Because
f
is not continuous at 0 by qn 6.20,
f
is not differentiable at 0 by
qn 12.
