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HW17solns

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

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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 δ > 0, we can set x =- δ/ 2 so that | x- | < δ , 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)....
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HW17solns - 1 h n i is an unbounded sequence with no...

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