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Su, Yung – Homework 10 – Due: Nov 6 2007, 3:00 am – Inst: Shinko Harper
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
IF
lim
n
→∞
a
n
= 7
,
determine the value, iF any, oF
lim
n
→∞
a
n
+8
.
1.
limit = 7
correct
2.
limit doesn’t exist
3.
limit =

1
4.
limit = 15
5.
limit =
7
8
Explanation:
To say that
lim
n
→∞
a
n
= 7
means that
a
n
gets as close as we please to 7
For all su±ciently large
n
. But then
a
n
+8
gets
as close as we please to 7 For all su±ciently
large
n
. Consequently,
lim
n
→∞
a
n
+8
= 7
.
keywords: sequence, limit, properties oF limits
002
(part 1 oF 1) 10 points
Determine iF the sequence
{
a
n
}
converges
when
a
n
=
1
n
ln
µ
4
2
n
+ 3
¶
,
and iF it does, fnd its limit.
1.
limit =

ln2
2.
limit = ln
4
5
3.
the sequence diverges
4.
limit = 0
correct
5.
limit = ln2
Explanation:
AFter division by
n
we see that
4
2
n
+ 3
=
4
n
2 +
3
n
,
so by properties oF logs,
a
n
=
1
n
ln
4
n

1
n
ln
µ
2 +
3
n
¶
.
But by known limits (or use L’Hospital),
1
n
ln
4
n
,
1
n
ln
µ
2 +
3
n
¶
→
0
as
n
→ ∞
. Consequently, the sequence
{
a
n
}
converges and has
limit = 0
.
keywords: limit, sequence, log Function,
003
(part 1 oF 1) 10 points
²ind a Formula For the general term
a
n
oF
the sequence
n
1
,

5
3
,
25
9
,

125
27
, ...
o
assuming that the pattern oF the frst Few
terms continues.
1.
a
n
=

‡
3
2
·
n
2.
a
n
=

‡
5
3
·
n
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2
3.
a
n
=
‡

3
2
·
n

1
4.
a
n
=
‡

5
3
·
n

1
correct
5.
a
n
=
‡

3
5
·
n

1
6.
a
n
=

‡
3
5
·
n
Explanation:
In the sequence
n
1
,

5
3
,
25
9
,

125
27
, ...
o
each term is

5
3
times the preceeding one,
i.e.
,
a
n
=
‡

5
3
·
a
n

1
.
Consequently,
a
n
=
‡

5
3
·
n

1
since
a
1
= 1.
keywords: sequence, exponential
004
(part 1 of 1) 10 points
Determine if the sequence
{
a
n
}
converges,
and if it does, Fnd its limit when
a
n
=
6
n
5

3
n
3
+ 5
5
n
4
+ 4
n
2
+ 4
.
1.
the sequence diverges
correct
2.
limit =
6
5
3.
limit =
5
4
4.
limit =

3
4
5.
limit = 0
Explanation:
After division by
n
4
we see that
a
n
=
6
n

3
n
+
5
n
4
5 +
4
n
2
+
4
n
4
.
Now
3
n
,
5
n
4
,
4
n
2
,
4
n
4
→
0
as
n
→ ∞
; in particular, the denominator
converges and has limit 5
6
= 0.
Thus by
properties of limits
{
a
n
}
diverges
since the sequence
{
6
n
}
diverges.
keywords:
005
(part 1 of 1) 10 points
Determine whether the sequence
{
a
n
}
con
verges or diverges when
a
n
=
7
n
2
7
n
+ 2

n
2
+ 3
n
+ 1
,
and if it does, Fnd its limit
1.
the sequence diverges
2.
limit =
5
14
3.
limit = 0
4.
limit =
5
21
5.
limit =
5
7
correct
Explanation:
After bringing the two terms to a common
denominator we see that
a
n
=
7
n
3
+ 7
n
2

(7
n
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 Spring '09
 Harper
 Calculus

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