1873_solutions

To show that no negative number can be a partial

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Unformatted text preview: though the sequence a n has a partial limit a that is less than a partial limit b of the sequence bn . We define a n  2  1 n and b n  3  1 n for every positive integer n. We see that a n  b n for each n and that 3 is a partial limit of a n and that 2 is a partial limit of b n . Some Exercises on the Algebraic Rules for Limits 1. Write out proofs of those cases of this theorem that were not proved above. We shall show just two more cases here. Proof of Part 3 when x  0 and y  Ý. Since x Ý  Ý, we need to show that x n y n Ý. Now since xn x  0 and y n Ý as n Ý, it follows from a case we have already considered that x n y n Ý as n Ý. From this fact and part 2 of the theorem we conclude that x n y n Ý. 139 Alternatively we can prove this case directly: We begin by choosing a number q such that x  q  0. (For example, one may define q  x/2. ) x To show that x n y n q 0 Ý, suppose that w is any real number. In view of the theorem of infinite limits we need to show that x n y n  w for all sufficiently large integers n. Using the fact that the interval Ý, q is a neighborhood of x and the fact that x n x we choose an integer N 1 such that x n  q for all integers n N 1 . Now we choose an integer N 2 such that y n  |w/q | for every integer n N 2 and we define N to be the larger of the two numbers N 1 and N 2 . Then for every integer n N we have x n y n  qy n  q w  |w | w q and the proof is complete. Proof of Part 4 when x  Ý and y is a positive real number Since Ý/y  Ý we need to prove that x n /y n Ý as n Ý. Since xn  x 1 ny yn n for each n and since 1/y n 1/y as n Ý the desired result follows from part 3 of the theorem. Alternatively, this part of the theorem can be proved directly. 2. Given that x n and y n are sequences of real numbers, that x n converges to a number x and that a real number y is a partial limit of the sequence 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 number x we choose an integer N such that the inequality |x n 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 |y n y |  2 holds. For each of these infinitely many integers we have |x n  y n x  y |  |x n x  y n y | |x n x |  |y n y|  2  2 . 3. State and prove some analogues of this exercise for subtraction, multiplication and division. 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 t...
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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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