1873_solutions

Of the following assertions is true or false if it is

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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.

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