(
x
2

9)
dx
= [9
x

x
3
/
3]
3

2
+ [
x
3
/
3

9
x
]
4
3
= ((27

27
/
3)

(

18

(

8)
/
3)) + ((64
/
3

36)

(27
/
3

27))
= 110
/
3
(b)
Z
2
π/
3

π/
2
cos(2
t
)
dt
=
1
2
Z
4
π/
3

π
cos(
t
)
dt
= [
1
2
sin(
t
)]
4
π/
3

π
=
1
2
(sin(4
π/
3)

sin(
π
)) =
1
2
(

√
3
/
2

0) =

√
3
4
In the first step we used the expansioncontraction properties of the integral.
1
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3. Give an ‘
ε

δ
’ proof of the statement
lim
x
→
1
√
x
+ 2 = 1.
Fix
ε >
0, we need to find
δ >
0 so that 0
<

x
+ 1

< δ
implies

√
x
+ 2

1

< ε
. This
inequality is equivalent to 1

ε <
√
x
+ 2
<
1 +
ε
. We may assume that
ε <
1 and then
the previous inequality is equivalent to
(1

ε
)
2

1
< x
+ 1
<
(1 +
ε
)
2

1
ε
2

2
ε < x
+ 1
< ε
2
+ 2
ε
(
*
)
Choosing
δ
= 2
ε

ε
2
=
ε
(2

ε
) (which is positive if
ε <
1) will get the job done.
Indeed, if

x
+1

< δ
then
ε
2

2
ε < x
+1
<
2
ε

ε
2
< ε
2
+2
ε
which means that inequality
(*) holds. But then

√
x
+ 2

1

< ε
holds as well which is what we needed.
4. Find the following limits. (You can use anything we proved in class about limits, but you
need to explain your steps!)
(a)
lim
x
→
0
sin(5
x
)
√
1 +
x

√
1

x
= lim
x
→
0
sin(5
x
)(
√
1 +
x
+
√
1

x
)
(
√
1 +
x

√
1

x
)(
√
1 +
x
+
√
1

x
)
= lim
x
→
0
sin(5
x
)(
√
1 +
x
+
√
1

x
)
2
x
= lim
x
→
0
sin(5
x
)
5
x
(
√
1 +
x
+
√
1

x
)
2
/
5
The function
(
√
1+
x
+
√
1

x
)
2
/
5
is continuous at zero so its limit is
(
√
1+0+
√
1

0)
2
/
5
= 5. The
limit of
sin(5
x
)
5
x
is the same as
sin(
x
)
x
as
x
→
0 which is 1. Using the fact that the limit
of products is the product of limits we get that
lim
x
→
0
sin(5
x
)
√
1 +
x

√
1
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 Fall '08
 Staff
 Math, Calculus, Derivative, Sin, lim, Continuous function

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