MAA 4200: Homework 5 Solutions
June 29, 2009
[3.2] We must show that lim
x
→
3
f
(
x
) =

12.
Preliminary Work: For
x
6
=

3,

f
(
x
) + 12

=

2(
x

3) + 12

=

2
x
+ 6

= 2

x
+ 3

.
Proof:
Let
>
0.
Choose
δ
=
/
2.
Then

f
(
x
) + 12

= 2

x
+ 3

<
2
δ
=
for all 0
<

x
+ 3

< δ
.
By Theorem 3
.
1,
f
is continuous at
x
=

3.
[3.6] Prove that
√
x
: [0
,
∞
) is continuous.
Proof
: First consider
x
0
= 0.
Let
>
0.
Choose
δ
=
2
.
Then for all
x
∈
D
with

x

< δ
we have,

√
x

=

x

1
/
2
< δ
1
/
2
=
.
Now we consider
x
0
>
0.
Let
>
0.
Choose
δ
=
√
x
0
.
Then for all
x
∈
D
with

x

x
0

< δ
we have,

√
x

√
x
0

=
1
√
x
+
√
x
0

x

x
0
 ≤
1
√
x
0

x

x
0

<
1
√
x
0
δ
=
.
[3.9] As proved in problem 2
.
6 in previous homework, lim
x
→
0
x
sin(1
/x
) = 0, so defining
f
(0) = 0 will make
f
continuous at 0 by Theorem 3
.
1.
[3.11] In problem 2
.
14 in previous homework, we proved that lim
x
→
1
f
(
x
) does not exist,
hence
f
is discontinuous at
x
= 1 by Theorem 3
.
1.
In problem 2
.
14 in previuos homework, we proved that lim
x
→
2
f
(
x
) = 16 and
f
(2) =
8
·
2 = 16 since 2 is rational. Therefore
f
is continuous at
x
= 2 by Theorem 3
.
1.
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 Spring '07
 thieme
 Continuous function, Odd, previous homework

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