Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. h n i is an unbounded sequence with no bounded, and hence no convergent, sub- sequence. h a n i defined by a n = n if n is odd 0 if n is even is an unbounded sequence that has h i as a convergent subsequence. 2. (a) Let h a n i = n i =1 1 /n . (b) Let h a n i = n i =1 n . (c) Let h a n i = n L . 3. Let h a n i = h 1 , 1 , 2 , 1 , 2 , 3 , 1 , 2 , 3 , 4 ,..., 1 , 2 ,...,k, 1 , 2 ,...,k + 1 ,... i . Then h n i is a subsequence of h a n i for every n N . 4. (a) Let = 1. Then if > 0, we can set x =- / 2 so that | x- | < , but | f ( x )- f (0) | = | (- 1)- 1 | = 2 > 1 = . Thus f is not continuous at 0. (b) Let a n =- 1 /n . Then a n 0, but f ( a n ) =- 1 6 1 = f (0). Hence f is not continuous at 0. 5. This is true. Set f ( x ) = x 2 3+ x 7 . First we notice that if 3 + x 7 = 0, then x < 0, so f is defined on [1 , ). Since f is a rational function, this means f is continuous on [1 , ). Now f (1) = 6 / 4 = 3 / 2 > 1 > 9 / 131 = f (2). By the Intermediate(2)....
View Full Document

This document was uploaded on 03/25/2011.

Page1 / 2

HW17solns - 1. h n i is an unbounded sequence with no...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online