Math 119
Week 2
Solutions
1
1. Guessing
δ
For given
>
0, we want to find
δ >
0 such that

(2
x
+ 3)

5

<
whenever

x

1

< δ
⇒ 
2(
x

1)
<
whenever

x

1

< δ
⇒ 
x

1

<
2
whenever

x

1

< δ
Thus, we can choose
δ
=
2
Showing that this
δ
works
∀
>
0
∃
δ
=
2
such that

(2
x
+ 3)

5

<
whenever

x

1

< δ
=
2

(2
x
+ 3)

5

=

2
x

2

= 2

x

1

<
2
.
2
=
2. Guessing
δ
For given
M
, we want to find
δ >
0 such that

1
(
x
+ 3)
4

> M
whenever

x

(

3)

< δ
⇒ 
(
x
+ 3)
4

<
1
M
whenever

x
+ 3

< δ
⇒ 
(
x
+ 3)

<
4
r
1
M
whenever

x
+ 3

< δ
so, we can choose
δ
=
4
r
1
M
Showing that this
δ
works
∀
>
0
∃
δ
=
4
r
1
M
such that

1
(
x
+ 3)
4

> M
whenever

x

(

3)

< δ
=
4
q
1
M

1
(
x
+ 3)
4

>
1
δ
4
=
1
(
4
q
1
M
)
4
=
M
3. We want that
f
(
x
) is continuous on (
∞
,
∞
). Obviously
f
(
x
) is continu
ous on (
∞
,
3) and (3
,
∞
), because
f
(
x
) is defined as a polinomial on the
ray (
∞
,
3) or (3
,
∞
). Thus, it is enough that
lim
x
→
3
+
f
(
x
) =
lim
x
→
3

f
(
x
) =
f
(3) to make it continuous.
lim
x
→
3
+
f
(
x
) =
lim
x
→
3
+
(
cx
2

1) = 9
c

1
lim
x
→
3

f
(
x
) =
lim
x
→
3

(
cx
+ 1) = 3
c
+ 1
f
(3) = 3
c
+ 1
=
⇒
9
c

1 = 3
c
+ 1
⇒
c
=
1
3
4. [

4
,

2)
,
(

2
,
2)
,
[2
,
4)
,
(4
,
6) and (6
,
8)
1
Please email all corrections and suggestions to these solutions to [email protected] All
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 Fall '08
 muhiddinuğuz
 Math, Calculus, Geometry, 4 M, 4 1 M, 119 Week, [email protected]

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