1873_solutions

In the interval v prove that x lmsup x n n the

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Unformatted text preview: subsequence in the set R S. Such a subsequence cannot be frequently in S. 5. If x n is a sequence of real numbers and S R, and if x n is not eventually in S then x n has a subsequence that is eventually in R S. Since a sequence that is not eventually in S must be frequently in R S, the desired result follows at once from a theorem on subsequences. Some Exercises on Limits and Partial Limits 1. Given that xn  3  1 n for each positive integer n, prove that 3 is a limit of x n . We need to prove that for every number  0 the sequence x n will eventually be in the interval 3 , 3  . So we start: Suppose that  0. Before we go any further we need to ask ourselves what it means to say that x n 3 , 3  . We observe that the inequality 3  x n  3  is equivalent to the assertion that 3  3 1  3 n which holds when 1/n  . This tells how how to continue: Using the fact that the number 1/ is not an upper bound of the set Z of integers we choose an integer N such that N  1/ , in other words, 1. N Then, whenever n N we have 3 1  3 3  3  3 1 n N and so we have shown that x n is eventually in 3 , 3  . 2. Given that xn  3  2 n for each positive integer n, prove that 3 is a limit of x n . We need to prove that for every number  0 the sequence x n will eventually be in the interval 3 , 3  . So we start: Suppose that  0. This time we make the observation that the inequality 3  3 2  3 n holds when 2/n  . Using the fact that the number 2/ is not an upper bound of the set Z of integers we choose an integer N such that N  2/ , in other words, 2. N Then, whenever n N we have 156  3  3 2 n is eventually in 3 3 and so we have shown that x n 3 2  3 N ,3  . 3. Given that x n  1/n for each positive integer n and that x 0, prove that x is not a partial limit of x n . Solution: In the event that x  0, the interval Ý, 0 is a neighborhood of x and it is clear that x n fails to be frequently in this neighborhood. Therefore no negative number can be a partial limit of x n . Suppose now that x  0. The interval x/2, Ý is a neighborhood of x x 2 0 x and the condition x n x/2, Ý must fail to hold whenever n  2/x. Therefore the sequence x n cannot be frequently in the interval x/2, Ý and the number x cannot be a partial limit of x n . 4. Given that 1 n n 3 if n is a multiple of 3 xn  0 if n is one more than a multiple of 3 , 4 if n is two more than a multiple of 3 Prove that the partial limits of x n are Ý, Ý, 0 and 4. Since x n is unbounded both above and below, it follows from the discussion of infinite partial limits we saw earlier that both Ý and Ý are partial limits of x n . Since the equation x n  0 holds for infinitely many values of n we know that x n is frequently in every neighborhood of 0 and so 0 is a partial limit of x n . In the same way we can see that 4 is a partial limit of x n . Now we need to explain why any real number other than 0 and 4 must fail to be a partial limit of x n . Suppose that x R 0, 4 . In the event that...
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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.

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