Analysis II Tutorial
Solution sketches
Alexandre Tomberg
January 13, 2011
1
Let (
x
n
) be a bounded sequence and let
L
be the set of its limit points. Let
(
a
k
) be a sequence in
L
and suppose that
a
= lim
k
→∞
a
k
exists.
Then
a
∈
L
.
Proof.
To show
a
∈
L
, it is enough to find a subsequence (
x
n
j
) of (
x
n
), s.t.
lim
j
→∞
x
n
j
=
a.
As
a
k
∈
L
,
∃
subsequence (
x
n
(
k
)
j
), s.t.
a
k
= lim
j
→∞
x
n
(
k
)
j
. Thus, for each
k
,
we an find an
x
n
k
s.t.

x
n
k

a
k

<
1
k
and
n
k
> n
k

1
.
Then, as

x
n
k

a

=

x
n
k

a
k
+
a
k

a
 ≤ 
x
n
k

a
k

+

a
k

a

<
1
k
+

a
k

a

,
taking
k
→ ∞
, we get that lim
k
→∞
x
n
k
=
a
.
1
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2
Find the limsup and lininf of the following sequence
x
n
=
n
2
1 +
n
2
cos
2
πn
3
.
Solution.
As
∀
n
,

1
2
= cos (2
π/
3) = cos
2
π
3
k

2
3
= cos
2
π
3
k

1
3
<
cos(2
πk
) = 1
.
Clearly, the subsequences
{
x
3
k

2
, x
3
k

1
}
and
{
x
3
k
}
converge, so that we get
lim inf
n
→∞
x
n
=
! lim
k
→∞
x
3
k

2
= lim
k
→∞
(3
k

2)
2
1 + (3
k

2)
2
·

1
2
=

1
2
,
lim sup
n
→∞
x
n
= lim
k
→∞
x
3
k
= lim
k
→∞
(3
k
)
2
1 + (3
k
)
2
= 1
.
Note that all other convergent subsequences must be contained in the two
subsequences we considered.
3
Construct a real sequence (
x
n
), whose set of
L
of limit points is s.t.
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 Spring '09
 Topology, Limit of a sequence, Xn, subsequence, lim xnj

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