48. continued
(b)
lim
x
→
2
1
1
f
(
x
)
5
0; lim
x
→
2
1
2
f
(
x
)
5
0
(c)
Yes. The limit is 0.
49. (a)
[
2
2
p
, 2
p
] by [
2
2, 2]
(b)
(
2
2
p
, 0)
<
(0, 2
p
)
(c)
c
5
2
p
(d)
c
5 2
2
p
50. (a)
[
2
p
,
p
] by [
2
3, 3]
(b)
1
2
p
,
}
p
2
}
2
<
1
}
p
2
}
,
p
2
(c)
c
5
p
(d)
c
5 2
p
51. (a)
[
2
2, 4] by [
2
1, 3]
(b)
(0, 1)
<
(1, 2)
(c)
c
5
2
(d)
c
5
0
52. (a)
[
2
4.7, 4.7] by [
2
3.1, 3.1]
(b)
(
2‘
,
2
1)
<
(
2
1, 1)
<
(1,
‘
)
(c)
None
(d)
None
53.
[
2
4.7, 4.7] by [
2
3.1, 3.1]
lim
x
→
0
(
x
sin
x
)
5
0
Confirm using the Sandwich Theorem, with
g
(
x
)
5 2
)
x
)
and
h
(
x
)
5
)
x
)
.
)
x
sin
x
)
5
)
x
)
?
)
sin
x
)
#
)
x
)
?
1
5
)
x
)
2
)
x
)
#
x
sin
x
#
)
x
)
Because lim
x
→
0
(
2
)
x
)
)
5
lim
x
→
0
)
x
)
5
0, the Sandwich Theorem
gives lim
x
→
0
(
x
sin
x
)
5
0.
54.
[
2
4.7, 4.7] by [
2
5, 5]
lim
x
→
0
(
x
2
sin
x
)
5
0
Confirm using the Sandwich Theorem, with
g
(
x
) =
2
x
2
and
h
(
x
)
5
x
2
.
)
x
2
sin
x
)
5
)
x
2
)
?
)
sin
x
)
#
)
x
2
)
?
1
5
x
2
.
2
x
2
#
x
2
sin
x
#
x
2
Because lim
x
→
0
(
2
x
2
)
5
lim
x
→
0
x
2
5
0, the Sandwich Theorem
gives lim
x
→
0
(
x
2
sin
x
)
5
0
55.
[
2
0.5, 0.5] by [
2
0.25, 0.25]
lim
x
→
0
1
x
2
sin
}
x
1
2
}
2
5
0
Confirm using the Sandwich Theorem, with
g
(
x
)
5 2
x
2
and
h
(
x
)
5
x
2
.
)
x
2
sin
}
x
1
2
}
)
5
)
x
2
)
?
)
sin
}
x
1
2
}
)
#
)
x
2
)
?
1
5
x
2
.
2
x
2
#
x
2
sin
}
x
1
2
}
#
x
2
Because lim
x
→
0
(
2
x
2
)
5
lim
x
→
0
x
2
5
0, the Sandwich Theorem
give lim
x
→
0
1
x
2
sin
}
x
1
2
}
2
5
0.
56.
[
2
0.5, 0.5] by [
2
0.25, 0.25]
lim
x
→
0
1
x
2
cos
}
x
1
2
}
2
5
0
Confirm using the Sandwich Theorem, with
g
(
x
)
5 2
x
2
and
h
(
x
)
5
x
2
.
)
x
2
cos
}
x
1
2
}
)
5
)
x
2
)
?
)
cos
}
x
1
2
}
)
#
)
x
2
)
?
1
5
x
2
.
2
x
2
#
x
2
cos
}
x
1
2
}
#
x
2
Because lim
x
→
0
(
2
x
2
)
5
lim
x
→
0
x
2
5
0, the Sandwich Theorem
give lim
x
→
0
1
x
2
cos
}
x
1
2
}
2
5
0.
57. (a)
In three seconds, the ball falls 4.9(3)
2
5
44.1 m, so its
average speed is
}
44
3
.1
}
5
14.7 m/sec.
46
Section 2.1