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Unformatted text preview: and y n x n x n y n for each n the present result follows at once from
Exercise 2.
11. Suppose that x n and y n are sequences of real numbers, that x n y n
a partial limit of at least one of the sequences x n and y n . Prove that
xn
1
yn
as n 0 and that the number 0 fails to be Ý. Solution: From Exercise 10 we know that x n and y n have the same sets of partial limits and we
know, therefore, that 0 is not a partial limit of either of these two sequences.
Using the fact that 0 is not a partial limit of y n , choose an integer N 1 and a number 0 such that the
inequality
y n 
holds whenever n N 1 . For every n N 1 we see that
x n y n 
xn 1 xn yn
yn
yn
and so the fact that
xn
1
yn
as n Ý follows from the sandwich theorem. 12. Give an example to show that the requirement in this exercise that 0 not be a partial limit of at least one of
the two sequences is really needed.
We define x n 2/n and y n 1/n for each n and observe that, even though x n y n 0 as n Ý,
x n /y n 2 as n Ý.
13. Suppose that x n and y n are sequences of real numbers, that x n /y n 1 and that at least one of the
sequences x n and y n is bounded. Prove that x n y n 0. Give an example to show that the conclusion
x n y n 0 can fail if both x n and y n are unbounded. Hint: Make the observation that both of the sequences x n and y n must be bounded. Explain
carefully how you arrive at this conclusion. Now make use of the fact that
x
x n y n  y n  y n 1
n
for every n.
14. Suppose that x n and y n are sequences of real numbers, that y n 1 and that for each n we have
z n x n y n . Prove that the sequences x n and z n have the same set of partial limits.
It follows from the part of Exercise 3 for which a solution is provided above that if x is an partial limit
of x n then the number x 1 x is a partial limit of the sequence z n . At the same time, since
x n z n /y n for all sufficiently large n we know that if x is any partial limit of z n then the number
x/1 x is a partial limit of the sequence x n . 142 Exercises on Sequences and the Topology of R
1. Prove that a set S of real numbers is unbounded below if and only if there exists a sequence x n in S such
that x n
Ý. Solution: We want to show that, for a given set S of real numbers, the following conditions are
equivalent:
a. The set S is unbounded below.
b. There exists a sequence x n in the set S such that x n Ý as n Ý. To show that condition b implies condition a, assume that condition b holds and choose a sequence x n in
S such that x n
Ý as n Ý. If w is any real number then it follows at once from the fact that the
sequence x n is eventually in the interval Ý, w that w is not a lower bound of S. Therefore S is not
bounded above.
Now to show that condition a implies condition b, assume that condition a holds. For each positive integer
n we use the fact that the number n is not a lower bound of S to choose a number that we shall call x n
such that x n n. The sequence x n that we have made in this wa...
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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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