# 9 - SOLUTIONS FOR HOMEWORK 4 9.1(b lim 3 n 7 6 n 5 = lim 3...

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Unformatted text preview: SOLUTIONS FOR HOMEWORK 4 9.1. (b) lim 3 n + 7 6 n- 5 = lim 3 + 7 /n 6- 5 /n = lim(3 + 7 /n ) lim(6- 5 /n ) = 3 6 = 1 2 . Here, we use the fact that, by Basic Example 9.7(a), lim(3 + 7 /n ) = 3 + 7 lim n − 1 = 3, and similarly, lim(6- 5 /n ) = 6. 9.2. (b) Note first that lim(3 y n- x n ) = lim(3 y n )- lim x n = 3 lim y n- lim x n = 3 · 7- 3 = 18 , and lim y 2 n = lim( y n · y n ) = lim y n · lim y n = (lim y n ) 2 = 7 2 = 49 . Therefore, lim 3 y n- x n y 2 n = lim(3 y n- x n ) lim y 2 n = 18 49 . 9.4. (b) Suppose s = lim s n . Letting s ′ n = s n +1 , we see that lim s ′ n = s . On the other hand, lim √ s n + 1 = radicalbig lim( s n + 1) = radicalbig lim s n + 1 = √ s + 1 . Therefore, passing to the limit on both sides of the identity s n +1 = √ s n + 1, we obtain: s = √ s + 1. Taking the square of both sides, we see that s is a solution to the quadratic equation s 2 = s + 1. However, s n ≥ 0 for every n , hence s ≥ 0 by Exercise 8.9. Therefore, s = ( √ 5 + 1) / 2. 9.9. (a) We have to show that for any A ∈ R there exists N s.t. t n > A for any n > N . We know that there exists a K s.t. s n > A for n > K . Then t n ≥ s n > A for n > N = max { N ,K } . (b) Similarly to (a), we have to show that, for any A ∈ R , there exists N s.t. s n < A for any n > N . There exists K s.t. t n < A for any n > K . Then s n < A for n > max { K,N } ....
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9 - SOLUTIONS FOR HOMEWORK 4 9.1(b lim 3 n 7 6 n 5 = lim 3...

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