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Unformatted text preview: 0 x 4, the fact that x is not a partial limit of x n follows from the fact that x n is
not frequently (or ever) in the interval 0, 4 which is a neighborhood of x.
Now suppose that x 0. x 1 x 0 In order to show that x is not a partial limit of x n we shall make the observation that x n is not
frequently in the interval x 1, 0 which is a neighborhood of x. In fact, the inequality
x 1 xn 0
can hold only when n is odd and
x 1 n3
31
which is equivalent to saying that n
x. Since there are only finitely many such positive
integers n we conclude that x n is not frequently in the interval x 1, 0
Finally we must consider the case x 4. x 4 x1 In this case we observe that there can be only finitely many positive integers n for which
4 n3 x 1
and so, once again, x can’t be a partial limit of x n .
5. Give an example of a sequence of real numbers whose set of partial limits is the set 1 Þ 4, 5 . Hint: For each positive integer n, if n can be written in the form
n 2m3k
for some positive integers m and k and if
4 157 m
k 5 then we define
xn m .
k
In all other cases we define x n 1. Observe that the range of the sequence x n is the set
1 Þ Q 4, 5
and then show that the set of partial limits of x n is 1 Þ 4, 5 .
Since the equation x n 1 holds for infinitely many positive integers n the number 1 must be a
partial limit of x n . To see that every number in the interval 4, 5 must be a partial limit of x n ,
suppose that x
4, 5 and suppose that 0. Since the interval x , x must contain infinitely
many members of the set Q 4, 5 we know that the condition x n
x , x must hold for
infinitely many positive integers n and so x must be a partial limit of x n .
Finally we observe that if x R
1 Þ 4, 5 then the open set R
1 Þ 4, 5 is a
neighborhood of x and that x n fails to be frequently (or ever) in R
1 Þ 4, 5 and so x can’t be
a partial limit of x n .
6. Given that
x n 3 2n
5n
for every positive integer n, prove that x n 2 as n Ý.
We begin by observing that if n is a positive integer then
3 2n 2 7
5n
5n
Now suppose that 0. The inequality
3 2n 2
5n
says that
7
5n
which holds when
5n 1
7
in other words
n 7 5
With these inequalities in mind we choose an integer N such that
N 7 5
and we observe that whenever n is an integer and n N we have
3 2n 2 .
5n
7. Given that
1
2n
1
n 2 1 xn if n is even
if n is odd prove that x n 0 as n Ý.
We observe that if n is a positive integer then
1.
n
0. Choose an integer N such that N 1/ . We observe that whenever n
1.
x n 0  x n 1
n
N
0 Now suppose that xn 8. Suppose that x n is a sequence of real numbers and that x
equivalent: 158 R. Prove that the following conditions are N, a. x n x as n Ý. b. For every number 5 ,x 5 . 0 the sequence x n is eventually in the interval x To prove that condition a implies condition b we assume that x n x as n Ý.
Suppose that 0 and, using the fact that x n x as n Ý, choose N such that the condition
xn
x ,x
w...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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