HW 4 Solutions
Math 115, Winter 2009, Prof. Yitzhak Katznelson
9.6 Let
x
1
= 1 and
x
n
+1
= 3
x
2
n
for
n
≥
1.
a) If
a
= lim
x
n
, then we have that
a
= lim
x
n
+1
= lim 3
x
2
n
=
3 lim(
x
n
·
x
n
) = 3(lim
x
n
)
2
= 3
a
2
, where the later steps use various
limit laws. Solving this equation yields two solutions; either
a
= 0 or
a
= 1
/
3.
b) Yes, the limit exists because
x
n
diverges to
∞
. To prove this, we
can first show by induction that
x
n
≥
3
n
−
1
; the basis for induction is
x
1
= 1, which is true. And if
x
n
≥
3
n
−
1
, then
x
n
+1
= 3
x
2
n
>
3
x
n
≥
3
·
3
n
−
1
= 3
n
, so that is the induction step. Now pick an
M >
0; we
want to find
N
such that
n > N
⇒
x
n
> M
. Let
N
= log
3
(
M
) + 1.
Then if
n > N
,
x
n
≥
3
n
−
1
>
3
N
−
1
= 3
log
3
(
M
)
=
M
.
squaresolid
c) There is no contradiction, because when we assume that lim
x
n
=
a
and do algebra to figure out what
a
is, we’re implicitly assuming that
a
is not
±∞
. As you may remember from calculus, the rules of algebra
don’t always work with
±∞
.
9.11 a) Suppose that lim
s
n
= +
∞
and inf
{
t
n
:
n
∈
N
}
>
−∞
; we
want to show that lim(
s
n
+
t
n
) = +
∞
. To do this, let
t
= inf
{
t
n
:
n
∈
N
}
; then
t >
−∞
.
Now pick
M >
0.
Since lim
s
n
=
∞
, there
exists
N
such that
n > N
implies that
s
n
>
max
{
M
−
t,
1
}
.
Then
s
n
> M
−
t
, so
s
n
+
t > M
.
But
t
= inf
{
t
n
}
, and so
t
n
≥
t
; thus
s
n
+
t
n
≥
s
n
+
t > M
whenever
n > N
. This is what we wanted.
squaresolid
(Aside:
The reason we use max
{
M
−
t,
1
}
instead of just
M
−
t
is that technically the definition 9.8 only guarantees the existence of
such an
N
when
M
−
t >
0, and
M
−
t
might not be positive. That
requirement is absolutely unnecessary, though, since the only thing we
really care about is large
M
, and any
N
that works for, say, 1 will also
work for any negative number. So we can replace “for
M >
0” in 9.8
with “for any
M
”, and similarly with
M <
0 in the second part of the
definition. I will not worry about whether
M >
0 again.)
b) Show that if lim
s
n
= +
∞
and lim
t
n
>
−∞
, then lim(
s
n
+
t
n
) =
+
∞
.
To do this, notice that
t
n
is a convergent sequence, so it is
bounded (Theorem 9.1). Then there is an
M
such that

t
n
 ≤
M
for
all
n
, so
t
n
≥ −
M
for all
n
, so inf
{
t
n
} ≥ −
M >
−∞
. Then part a)
immediately gives us what we want.
squaresolid
c) If lim
s
n
= +
∞
and
t
n
is bounded, then just as in b), inf
{
t
n
}
>
−∞
. Thus by part a), we are done immediately, and lim(
s
n
+
t
n
) =
+
∞
.
squaresolid
1
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9.12. Assume all
s
n
negationslash
= 0 and that the limit
L
= lim

s
n
+1
s
n

exists.
a) Show that if
L <
1, then lim
s
n
= 0.
First select
a
so that
L < a <
1; notice that, say, the average of
L
and 1 would work.
Now, lim

s
n
+1
s
n

=
L < a
, so let
ǫ
=
a
−
L
; then
there exists an
N
′
where
n > N
′
⇒ 
s
n
+1
s
n
 −
L

< ǫ
. In particular, if
we let
N
=
N
′
+ 1, then
n
≥
N
implies that

s
n
+1
s
n

< L
+
ǫ
=
a
, i.e.
that

s
n
+1

< a

s
n

. So certainly

s
N
+1

< a

s
N

.
Now we see that

s
N
+2

< a

s
N
+1

< a
2

s
N

. This pattern continues,
and we see that

s
N
+
k

< a
k

s
N

for any
k >
0.
(Technically this
requires induction, but this is obvious enough that you could just state
it and say it’s proved by induction).
Changing variables and letting
n
=
N
+
k
, we see that for
n > N
, we have

s
n

< a
n
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 Limit of a function, Limit of a sequence, subsequence

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