Math 3000
Fall 2020
Quiz 7
Note that for all
x >
0, one has
√
x
2
+ 1
>
√
x
2
=
x
. Therefore, one has
1
√
x
2
+ 1 +
x
≤
1
2
x
≤
1
x
=
1
x
.
We now let
1
x
<
which implies that
x >
1
/
. Now let
G
( ) = 1
/
. That is, for all
>
0, one can
find
G
( ) = 1
/
, such that for all
x > G
( ), one has

f
(
x
)

L

<
and hence
lim
x
→
+
∞
h
p
x
2
+ 1

x
i
=
lim
x
→
+
∞
1
√
x
2
+ 1 +
x
= 0
.
3. (4 marks) Find and classify the discontinuity of the following function
f
(
x
) =
1

x
1

x
3
.
ANS:
First of all, the domain of this function is 1

x
3
6
= 0 which leads to
x
6
= 1 as its domain.
We let
D
= (
∞
,
1)
∪
(1
,
∞
). So we check the discontinuity of point
x
= 1. Note that
L
1
=
lim
x
→
1

f
(
x
) =
lim
x
→
1

1

x
1

x
3
=
lim
x
→
1

1

x
(1

x
)(1 +
x
+
x
2
)
=
lim
x
→
1

1
1 +
x
+
x
2
=
1
3
,
L
2
=
lim
x
→
1
+
f
(
x
) =
lim
x
→
1
+
1

x
1

x
3
=
lim
x
→
1
+
1

x
(1

x
)(1 +
x
+
x
2
)
=
lim
x
→
1
+
1
1 +
x
+
x
2
=
1
3
.
It follows that
L
=
L
1
=
L
2
=
1
3
, however,
f
(
x
) is undefined at
x
= 1. Hence
x
= 1 is the removable
discontinuity.
4. (6 marks) Prove or disprove the uniform continuity of
f
(
x
) =
x
3

6
x

1 on
I
= (0
,
5). Prove that
f
(
x
) = 0 must have one real solution.
ANS:
For any
>
0, for all
x, y
∈
I
, we consider the following inequality

f
(
x
)

f
(
y
)

=

x
3

6
x

1

y
3
+ 6
y
+ 1

=

x
3

y
3

6(
x

y
)

=

x

y

x
2
+
xy
+
y
2

6

<
.
In fact, for all
x, y
∈
I
, one has

x
2
+
xy
+
y
2

6
 ≤
x
2
+

xy

+
y
2
+ 6
≤
81 =
M.
We now let
M

x

y

<
⇔ 
x

y

<
/M.
Let
δ
=
/M
. Then, for any
>
0, there is
δ
=
/M
, such that for all
x, y
∈
I
with

x

y

< δ
, one
has

f
(
x
)

f
(
y
)

=

x
3

6
x

1

y
3
+ 6
y
+ 1

=

x

y

x
2
+
xy
+
y
2

6

<
.
Hence
f
(
x
) =
x
3

6
x

1 is uniformly continuous on
I
= (0
,
5).
Then we show
f
(
x
) = 0 must have one real solution.
Since
f
(
x
) is uniformly continuous on
I
= (0
,
5), it follows that
f
(
x
) is continuous on
I
= (0
,
5). Note that
lim
x
→
0
f
(
x
) =
f
(0) =

1
<
0
and
lim
x
→
5
f
(
x
) =
f
(5) = 94
>
0
.
One has
f
(
x
) is continuous on
I
0
= [0
,
5]. By Intermediate Value Theorem, there exists
c
∈
I
0
= [0
,
5]
such that
f
(
c
) = 0. Hence
f
(
x
) = 0 must have one real solution.
2