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Unformatted text preview: teger n 1, and then show that the sequence x 2n 1 is increasing and that the sequence x 2n
is decreasing and that these two sequences have the same limit 2 1. Solution: For every positive number x we define
Whenever 0 t x we see that g x 2x .
5 2x 173 g t 2x
5 2x 2t
5 2t xt
0
5 2x 5 2t
and so the function g is strictly increasing. Since x 1 x 3 we have g x 1 g x 3 from which we
deduce that x 3 x 5 . Continuing in this way we see that the sequence x 2n 1 is increasing. Since
x 2 x 4 we have g x 2 g x 4 from which deduce that x 4 x 5 . Continuing in this way we see that
the sequence x 2n is decreasing. Therefore the sequences x 2n 1 and x 2n are convergent. If we
write
x n Ý x 2n 1
lim
gx then it follows from the identity
x 2 n 1 2 x 2 n 1
5 2x 2n 1
that
x 2x
5 2x
and we can see that the only positive solution of this equation is 2
lim x
n Ý 2n 1 2 1. Thus 1 and we can see similarly that
lim x
n Ý 2n
c. Deduce that x n 2 1 as n 2 1. Ý. Exercises on Upper and Lower Limits
1. Prove that if x n is a sequence of real numbers then x n has a limit if and only if
lmsup x n lminf x n .
nÝ nÝ Solution: This exercises follows at once from an earlier theorem.
2. Prove that a sequence x n of real numbers is bounded above if and only if
lmsup x n Ý.
nÝ We know that a sequence is bounded above if and only if Ý is not a partial limit of the sequence.
There a sequence is bounded above if and only if its largest partial limit is not Ý.
3. Suppose that x n is a sequence of real numbers.
a. Prove that if x is a partial limit of the sequence x n then the number x is a partial limit of the sequence
xn .
if Ý is a partial limit of a sequence x n then x n is unbounded above and so x n is
unbounded below, making Ý a partial limit of x n .
Now suppose that a real number x is a partial limit of a given sequence x n . To show that the
number x is a partial limit of the sequence x n , suppose that 0.
We know that there are infinitely many integers n for which
x
xn x
and for each of these integers n we have
x
xn x .
Therefore the sequence x n is frequently in the interval x , x and we conclude that
x is a partial limit of x n .
b. Prove that 174 lmsup x n
nÝ lminf x n .
nÝ Since lminf n Ý x n is a partial limit of x n we know from part a that lminf n Ý x n is a partial limit
of the sequence x n . Now given any partial limit q of x n , we know that q, being a partial
limit of x n , cannot be less than lminf n Ý x n . In other words, whenever q is a partial limit of
x n we have
q
lminf x n
nÝ which gives us
q
Therefore lminf n Ý xn lminf x n .
nÝ is the largest partial limit of the sequence xn . 4. Suppose that x n is a sequence of real numbers, that
x lmsup x n
nÝ and that u x v.
a. Prove that the sequence x n must be bounded above.
The given inequality u x v tells us that x Ý and so x n is bounded above.
b. Prove that the sequence x n must be frequently in the interval u, Ý .
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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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