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Unformatted text preview: 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 135 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:
a. x n x as n N, R. Prove that the following conditions are 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
will hold whenever n N. Then, whenever n N we have
x ,x
x 5 ,x 5 .
xn
Now to prove that condition b implies condition a we assume that condition b holds. To prove that
condition a holds, suppose that 0. Using the fact that /5 is a positive number, we choose an
integer N such that the condition
xn
holds whenever n ,x 5
5
5
N we have x N. Thus for every n
xn x 5 5 5 ,x 5 5 x ,x . 9. Prove that
n 2 3n 1
2n 2 n 4 1
2 as n Ý.
We begin by observing that if n is any positive integer then
5n 5 .
n 2 3n 1
5n 2
1
2
4n
4n 2
2n 2 n 4
4n 2 2n 8
Now suppose that 0. The inequality
n 2 3n 1
1
2
2n 2 n 4
will hold when
5
4n
which requires that
n 5
4
Choose an integer N such that N 5/ 4 and observe that, whenever n N we have
5
5.
n 2 3n 1
1
2
4n
4N
2n 2 n 4
10. For each positive integer n, if n can be written in the form 136 n 2m3k
where m and k are postive integers and 0
1 then we define
xn m .
k
Otherwise we define x n 0. Prove that the set of partial limits of the sequence x n is 0, 1 .
Since the range of x n is the set of all rational numbers in the interval 0, 1 , every neighborhood of
a number x in the interval 0, 1 must contain infinitely many members of the range of x n and
must, therefore, contain the number x n for infinitely many integers n. Thus every member of the
interval 0, 1 is a partial limit of x n .
If x is any number in...
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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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