SOLUTIONS TO HOMEWORK # 5
9.2
(a) lim(
x
n
+
y
n
) = lim
x
n
+ lim
y
n
= 3 + 7 = 10.
(b) lim
3
y
n

x
n
y
2
n
=
lim(3
y
n

x
n
)
lim
y
2
n
=
lim3
y
n

lim
x
n
(lim
y
n
)
2
=
3
·
7

3
7
2
=
18
49
.
9.3
lim
±
a
3
n
+ 4
a
n
b
2
n
+ 1
¶
=
lim(
a
3
n
+ 4
a
n
)
lim(
b
2
n
+ 1)
=
lim
a
3
n
+ lim 4
a
n
lim
b
2
n
+ lim1
=
(lim
a
n
)
3
+ 4 lim
a
n
(lim
b
n
)
2
+ 1
=
a
3
+ 4
a
b
2
+ 1
.
9.5
If lim
t
n
exists, then lim
t
n
+1
= lim
±
t
2
n
+ 2
2
t
n
¶
=
lim(
t
2
n
+ 2)
lim(2
t
n
)
=
(lim
t
n
)
2
+ 2
2lim
t
n
,
from the limit theorems. Now let lim
t
n
=
t
. Then lim
t
n
+1
=
t
since the sequences
(
t
n
) and (
t
n
+1
) diﬀer only by the term
t
1
. We have
t
=
t
2
+ 2
2
t
,
2
t
2
=
t
2
+ 2
, t
2
= 2 or
t
=
±
√
2
.
So which of these solutions is the limit? To answer this question we note that
all
terms of
t
n
are positive
. We prove this by Mathematical Induction.
Step 1. It is given that
t
1
= 1 so the base of induction holds (it is here where we
use the information
t
1
= 1 provided in the statement of the problem!).
Step 2. Assume
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 Spring '08
 OLSAVSKYGREGORM
 Limit, lim, lim tn

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