Problem Set 2
Calculus 1
Abdullah Khalid, Hassan Bukhari
November 23, 2010
Exercise 2.3
9.
We have
f
(
x
) =
√
x,x
o
= 1
,L
= 1
,±
=
1
4
.
We have the graph given in the book. As can be seen
L
+
±
= 1 +
1
4
=
5
4
and
L

±
=
1

1
4
=
3
4
. Consider that when
f
(
x
) lies in the interval (
3
4
,
5
4
),
x
is in the interval (
9
16
,
25
16
),
and
x
o
= 1 belongs to this interval. Now the lenght of (
9
16
,
1) is smaller than length of
(1
,
25
16
). So we let
δ
= 1

9
16
=
7
16
.
We can also solve this algebrically. Let’s start with,

f
(
x
)

L

< ±
Plug in the values of everything and manipulate,

√
x

1

<
1
4

1
4
<
√
x

1
<
1
4
3
4
<
√
x <
5
4
9
16
< x <
25
16

7
16
< x

1
<
9
16
⇒ 
x

1

<
7
16
which is of the form,

x

x
o

<
7
16
Therefore, we have
δ
=
7
16
.
10.
We have
f
(
x
) = 2
√
x
+ 1
,x
o
= 3
,L
= 4
,±
= 0
.
2 =
1
5
.
Again we can see that
L
+
±
= 4 + 0
.
2 = 4
.
2 and
L

±
= 1

0
.
2 = 3
.
8. Consider that
1
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View Full Documentwhen f(x) lies in the interval (3
.
8
,
4
.
2), x is in the interval (2
.
61
,
3
.
41), and
x
o
= 3 belongs
to this interval. Now the lenght of (2
.
61
,
1) is smaller than length of (3
,
3
.
41). So we let
δ
= 3

2
.
61 = 0
.
39.
Solving algebrically, we proceed as before,

f
(
x
)

L

< ±

2
√
x
+ 1

4

<
1
5

1
5
<
2
√
x
+ 1

4
<
1
5

19
5
<
2
√
x
+ 1
<
21
5

19
10
<
√
x
+ 1
<
21
10

19
2
10
2
< x
+ 1
<
21
2
10
2
⇒ 
0
.
39
< x

3
<
0
.
41
⇒ 
x

3

<
0
.
39
which is of the form,

x

x
o
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 Spring '10
 RazaSuleman
 Calculus, Limit, Existence, Limit of a function, XO

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