1873_solutions

The interval prove that there exists an 0 is a

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Unformatted text preview: s for Limits 1. Suppose that x n and y n are sequences in R k , that x n x as n Ý and that y is a partial limit of y n . Prove that x  y is a partial limit of the sequence x n  y n . Suppose that  0. Using the fact that x n converges to the point x we choose an integer N such that the inequality xn x  2 holds whenever n N. Since y is a partial limit of the sequence y n and since there are only finitely many positive integers n  N there must be infinitely many integers n N for which the inequality yn y  2 holds. For each of these infinitely many integers we have xn  yn x  y  xn x  yn y x  yn xn y 2  2 . 2. Is it true that if x n and y n are sequences in R k and x is a partial limit of x n and that y is a partial limit of y n then x  y is a partial limit of the sequence x n  y n ? The answer is no; even in the metric space R. Define x n  y n  1 n for each n. We observe that the numbers 1 and 1 are partial limits of x n and y n respectively but that 1  1 fails to be a partial limit of x n  y n . 3. State and prove some analogues of Exercise 1 for differences and inner products in R k and for products and quotients in R. Suppose that x n and y n are sequences of numbers, that x n converges to a number x and that a real number y is a partial limit of y n . We shall prove that the number xy must be a partial limit of the sequence x n y n . Using the fact that x n is convergent, and therefore bounded, we choose a number p such that |x n |  p for every n. For each n we observe that |x n y n xy |  |x n y n x n y  x n y xy | |x n y n p|y n x n y |  |x n y y |  |y ||x n xy | x| Now, to show that xy is a partial limit of the sequence x n y n , suppose that  0. Using the fact that x n x as n Ý we choose an integer N such that the inequality |x n x |  2|y |  1 holds whenever n N. Since there are only finitely many positive integers n  N, and y is a partial limit of the sequence y n , there must be infinitely many integers n N for which the inequality |y n y |  2p holds. For each of these infinitely many integers n we have |x n y n xy |  |x n y n x n y  x n y xy | |x n y n p|y n p 2p 164 x n y |  |x n y y |  |y ||x n  |y | xy | x| 2|y |  1 . 4. Suppose that x n and y n are sequences in R 3 that converge respectively to points x and y. Prove that lim x y n  x y. nÝ n We write each point x n in the form xn  an, bn, cn and each point y n in the form yn  un, vn, wn and x  a, b, c and The given information tells us that as n w n w. Therefore yn  n Ý bnwn lim lim x nÝ n  bw cv, cu y  u, v, w . Ý we have a n a, b n cnvn, cnun anwn, anvn aw, av b, c n bu  a, b, c c, u n u, v n v and bnun u, v, w  x y. 5. Give an example of two divergent sequences x n and y n in R such that the sequence x n  y n is convergent. We define x n  1 n and y n  1 n 1 for each n. Note that the sequence x n  y n is the sequence with constant value 0. 6. Give an example of two sequences x n and y n in R such that x n 0 and y n Ý and a. x n y n 0 We...
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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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