fh2825 – Homework 10 – Radin – (58415)
1
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23
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001
10.0 points
Compute the value of
lim
n
→∞
6
a
n
b
n
5
a
n
−
6
b
n
when
lim
n
→∞
a
n
= 6
,
lim
n
→∞
b
n
=
−
2
.
1.
limit =
−
12
7
2.
limit =
12
7
3.
limit =
37
21
4.
limit =
−
37
21
5.
limit doesn’t exist
Explanation:
By properties of limits
lim
n
→
2
6
a
n
b
n
= 6 lim
n
→∞
a
n
lim
n
→∞
b
n
=
−
72
while
lim
n
→∞
(5
a
n
−
6
b
n
)
= 5 lim
n
→∞
a
n
−
6 lim
n
→∞
b
n
= 42
negationslash
= 0
.
Thus, by properties of limits again,
lim
n
→∞
6
a
n
b
n
5
a
n
−
6
b
n
=
−
12
7
.
002
10.0 points
If
lim
n
→ ∞
a
n
= 6
,
determine the value, if any, of
lim
n
→ ∞
a
n
−
8
.
1.
limit doesn’t exist
2.
limit = 6
3.
limit =
−
2
4.
limit = 14
5.
limit =
3
4
Explanation:
To say that
lim
n
→ ∞
a
n
= 6
means that
a
n
gets as close as we please to 6
for all sufficiently large
n
. But then
a
n
−
8
gets
as close as we please to 6 for all sufficiently
large
n
. Consequently,
lim
n
→ ∞
a
n
−
8
= 6
.
003
10.0 points
Find a formula for the general term
a
n
of
the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
1
,
−
5
4
,
25
16
,
−
125
64
, . . .
bracerightBig
,
assuming that the pattern of the first few
terms continues.
1.
a
n
=
−
parenleftBig
4
5
parenrightBig
n
2.
a
n
=
parenleftBig
−
6
5
parenrightBig
n
−
1
3.
a
n
=
−
parenleftBig
5
4
parenrightBig
n
4.
a
n
=
−
parenleftBig
6
5
parenrightBig
n
Changed with the DEMO VERSION of CADKAS PDFEditor (http://www.cadkas.com).
Changed with the DEMO VERSION of CADKAS PDFEditor (http://www.cadkas.com).
Changed with the DEMO VERSION of CADKAS PDFEditor (http://www.cadkas.com).
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fh2825 – Homework 10 – Radin – (58415)
2
5.
a
n
=
parenleftBig
−
5
4
parenrightBig
n
−
1
6.
a
n
=
parenleftBig
−
4
5
parenrightBig
n
−
1
Explanation:
By inspection, consecutive terms
a
n
−
1
and
a
n
in the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
1
,
−
5
4
,
25
16
,
−
125
64
, . . .
bracerightBig
have the property that
a
n
=
ra
n
−
1
=
parenleftBig
−
5
4
parenrightBig
a
n
−
1
.
Thus
a
n
=
ra
n
−
1
=
r
2
a
n
−
2
=
. . .
=
r
n
−
1
a
1
=
parenleftBig
−
5
4
parenrightBig
n
−
1
a
1
.
Consequently,
a
n
=
parenleftBig
−
5
4
parenrightBig
n
−
1
since
a
1
= 1.
keywords: sequence, common ratio
004
10.0 points
Determine if the sequence
{
a
n
}
converges,
and if it does, find its limit when
a
n
=
5
n
5
−
2
n
3
+ 4
5
n
4
+ 3
n
2
+ 5
.
1.
limit =
−
2
3
2.
limit = 1
3.
the sequence diverges
4.
limit = 0
5.
limit =
4
5
Explanation:
After division by
n
4
we see that
a
n
=
5
n
−
2
n
+
4
n
4
5 +
3
n
2
+
5
n
4
.
Now
2
n
,
4
n
4
,
3
n
2
,
5
n
4
−→
0
as
n
→ ∞
; in particular, the denominator
converges and has limit 5
negationslash
=
0.
Thus by
properties of limits
{
a
n
}
diverges
since the sequence
{
5
n
}
diverges.
005
10.0 points
Determine whether the sequence
{
a
n
}
con
verges or diverges when
a
n
=
8
n
2
8
n
+ 5
−
n
2
+ 6
n
+ 1
,
and if it does, find its limit
1.
limit =
3
16
2.
limit = 0
3.
the sequence diverges
4.
limit =
3
8
5.
limit =
1
8
Explanation:
After bringing the two terms to a common
denominator we see that
a
n
=
8
n
3
+ 8
n
2
−
(8
n
+ 5)
(
n
2
+ 6
)
(8
n
+ 5) (
n
+ 1)
=
3
n
2
−
48
n
−
30
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 Spring '08
 RAdin
 Squeeze Theorem, Limit, lim, Limit of a function, CADKAS PDFEditor, DEMO VERSION

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