But
parenleftBig
1 +
x
n
parenrightBig
n
−→
e
x

cadena (jc59484) – HW10 – lawn – (55930)
3
as
n
→
∞
.
Consequently,
{
a
n
}
converges
and has
limit =
(
e
−
1
)
−
6
=
e
6
.
006
10.0points
Determine whether the sequence
{
a
n
}
con-
verges or diverges when
a
n
= (
−
1)
n
parenleftbigg
6
n
+ 3
7
n
+ 7
parenrightbigg
,
and if it does, find its limit.
1.
sequence diverges
correct
2.
limit =
3
7
3.
limit =
±
6
7
4.
limit =
6
7
5.
limit = 0
Explanation:
After division,
6
n
+ 3
7
n
+ 7
=
6 +
3
n
7 +
7
n
.
Now
3
n
,
7
n
→
0 as
n
→ ∞
,
so
lim
n
→ ∞
6
n
+ 3
7
n
+ 7
=
6
7
negationslash
= 0
.
Thus as
n
→ ∞
, the values of
a
n
oscillate be-
tween values ever closer to
±
6
7
. Consequently,
the sequence diverges
.
007
10.0points
Determine if the sequence
{
a
n
}
converges,
when
a
n
= 4 cos
parenleftbigg
2
n π
+ 1
6
n
+ 7
parenrightbigg
,
and if it does, find its limit.
1.
limit = 2
√
3
2.
limit =
4
3
π
3.
limit = cos
1
7
4.
limit = 4 cos
1
7
5.
limit = 2
correct
6.
sequence does not converge
Explanation:
After division,
2
n π
+ 1
6
n
+ 7
=
2
π
+
1
n
6 +
7
n
.
But
1
n
,
7
n
→
0 as
n
→ ∞
,
so
lim
n
→ ∞
2
n π
+ 1
6
n
+ 7
=
π
3
.
Consequently, since cos
x
is continuous as a
function of
x
, the sequence
{
a
n
}
converges
and has
limit = 4 cos
π
3
= 2
.
008
10.0points
Determine if the sequence
{
a
n
}
converges
when
a
n
=
(2
n
+ 1)!
7
n
2
(2
n
−
1)!
,
and if it converges, find the limit.
1.
does not converge

cadena (jc59484) – HW10 – lawn – (55930)
4
2.
converges with limit =
7
2
3.
converges with limit =
4
7
correct
4.
converges with limit =
2
7
5.
converges with limit =
7
4
Explanation:
By definition,
m
! is the product
m
! = 1
·
2
·
3
·
. . .
·
m
of the first
m
positive integers.
When
m
=
2
n
−
1, therefore,
(2
n
−
1)! = 1
·
2
·
3
·
. . .
·
(2
n
−
1)
,
while
(2
n
+ 1)! = 1
·
2
·
3
·
. . .
·
(2
n
−
1)2
n
(2
n
+ 1)
.
when
m
= 2
n
+ 1. But then,
(2
n
+ 1)!
7
n
2
(2
n
−
1)!
=
2
n
(2
n
+ 1)
7
n
2
−→
4
7
as
n
→ ∞
. Consequently, the given sequence
converges with limit =
4
7
.
009
10.0points
Which of the following sequences converge?
A
.
braceleftbigg
5
e
n
2 +
e
n
bracerightbigg
B
.
braceleftbigg
e
n
+ 5
6
n
+ 4
bracerightbigg
1.
both
A
and
B
2.
B
only
3.
neither of them
4.
A
only
correct
Explanation:
A
. After simplification,
5
e
n
2 +
e
n
=
5
2
e
−
n
+ 1
.
Thus
5
e
n
2 +
e
n
−→
5
as
n
→ ∞
since
e
−
x
→
0 as
x
→ ∞
.
B
. To test for convergence of the sequence
braceleftbigg
e
n
+ 5
6
n
+ 4
bracerightbigg
we use L’Hospital’s Rule with
f
(
x
) =
e
x
+ 5
,
g
(
x
) = 6
x
+ 4
since
f
(
x
)
−→ ∞
,
g
(
x
)
−→ ∞
as
x
→ ∞
. Thus
lim
n
→ ∞
e
n
6
n
+ 4
=
lim
x
→ ∞
f
′
(
x
)
g
′
(
x
)
.
But
f
′
(
x
)
g
′
(
x
)
=
e
x
6
−→ ∞
as
n
→ ∞
.
Consequently,
only
A
converges
.
010
10.0points
Determine whether the sequence
{
a
n
}
con-
verges or diverges when
a
n
=
ln(5
n
6
)
ln(3
n
5
)
,
and if it converges, find the limit.
1.
diverges
2.
converges with limit
= 0

cadena (jc59484) – HW10 – lawn – (55930)
5
3.
converges with limit =
6
5
correct
4.
converges with limit
=
ln(5)
ln(3)
5.
converges with limit =
5
3
Explanation:
By properties of logs,
ln(5
n
6
) = ln(5) + 6 ln(
n
)
,
ln(3
n
5
) = ln(3) + 5 ln(
n
)
.

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- Fall '11
- Gramlich
- Accounting, Limit, lim