Math 3110 Homework 10 Solutions
Exercise 11.1.1
We can compute

f
(
x
)

f
(1)

=
x
1 +
x

1
1 + 1
=
2
x
2(1 +
x
)

1 +
x
2(1 +
x
)
=
x

1
2(
x
+ 1)
.
Given
>
0, let
δ
= min
{
1
,
2
}
. If

x

1

< δ
≤
1, then we have

x

1

<
1, from which
0
< x <
2, and so 1
< x
+ 1
<
3. Hence,

f
(
x
)

f
(1)

=
x

1
2(
x
+ 1)
<

x

1

2
<
δ
2
<
.
Exercise 11.4.4
By exercise 2.4.1, max(
a, b
) =
a
+
b
+

a

b

2
. If we let
a
=
f
(
x
) and
b
=
g
(
x
), this holds
for any particular value of
x
, so max(
f, g
)(
x
) =
f
(
x
) +
g
(
x
) +

f
(
x
)

g
(
x
)

2
. We know that
f
(
x
)

g
(
x
) is continuous by linearity. By question 11.4.3,

f
(
x
)

g
(
x
)

is continuous. By
linearity again (twice if you like), max(
f, g
)(
x
) =
f
(
x
) +
g
(
x
) +

f
(
x
)

g
(
x
)

2
is continous.
Similarly, min(
f, g
)(
x
) =
f
(
x
) +
g
(
x
)
 
f
(
x
)

g
(
x
)

2
is also continuous by linearity.
Alternatively, we could do this one directly from the definition of continuity. We show that
the functions continuous at an arbitrary point
x
=
x
0
. Let
>
0 be given. Since
f
and
g
are
continuous at
x
0
, there are
δ
1
, δ
2
>
0 such that if

x

x
0

< δ
1
, then

f
(
x
)

f
(
x
0
)

<
and if

x

x
0

< δ
2
, then

g
(
x
)

g
(
x
0
)

<
. Let
δ
= min
{
δ
1
, δ
2
}
. If

x

x
0

< δ
, then

x

x
0

< δ
1
and

x

x
0

< δ
2
, so

f
(
x
)

f
(
x
0
)

<
and

g
(
x
)

g
(
x
0
)

<
.
Assume without loss of generality that
f
(
x
0
)
≥
g
(
x
0
).
If

x

x
0

< δ
and
f
(
x
)
≥
g
(
x
),
then

max(
f, g
)(
x
)

max(
f, g
)(
x
0
)

=

f
(
x
)

f
(
x
0
)

<
, which is what we needed to show.
1
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Otherwise,
g
(
x
)
> f
(
x
), and we have

max(
f, g
)(
x
)

max(
f, g
)(
x
0
)

=

g
(
x
)

f
(
x
0
)

. We
compute
g
(
x
)

f
(
x
0
)
≥
g
(
x
)

g
(
x
0
)
>

. Furthermore,
g
(
x
)

f
(
x
0
)
≤
f
(
x
)

f
(
x
0
)
<
.
Therefore,

< g
(
x
)

f
(
x
0
)
<
, which gives

g
(
x
)

f
(
x
0
)

<
, and we are done.
For any particular value of
x
, max(
f
(
x
)
, g
(
x
)) and min(
f
(
x
)
, g
(
x
)) are just
f
(
x
) and
g
(
x
) in
some order. Hence, max(
f
(
x
)
, g
(
x
)) + min(
f
(
x
)
, g
(
x
)) =
f
(
x
) +
g
(
x
), from which we have
min(
f
(
x
)
, g
(
x
)) =
f
(
x
) +
g
(
x
)

max(
f
(
x
)
, g
(
x
)). Then by Theorem 11.4C, min(
f
(
x
)
, g
(
x
))
is continuous because it is a linear combination of continuous functions.
(We could have
repeated the reasoning to show that max(
f
(
x
)
, g
(
x
)) is continuous, but citing a theorem is
faster.)
Exercise 11.5.1
(a)
Since
x
is a real number, there is a sequence
{
a
n
}
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 '08
 RAMAKRISHNA
 Math, Ri, sequential continuity theorem, ln x−ln

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