# hw4sol - Math 3110 Homework 4 Solutions Exercise 5.1.4 an...

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Math 3110 Homework 4 Solutions Exercise 5.1.4 Claim. If a n b n L , b n 6 = 0 for any n , and b n 0 , then a n 0 . Proof. Since lim n →∞ a n b n = 0 and lim n →∞ b n = 0, then by the product theorem (5.1(2)), lim n →∞ ± a n b n · b n ² = L · 0 = 0 . But since a n = a n b n · b n , then lim n →∞ a n = lim n →∞ ± a n b n · b n ² , so we have shown that lim n →∞ a n = 0 . Exercise 5.2.1 Claim. lim n →∞ n + cos n n + 1 = 1 . Proof. Since - 1 cos n 1 for all n , then n - 1 n + 1 n + cos n n + 1 n + 1 n + 1 for all n . Re-writing the left and right hand sides, we have that for all n , r n n + 1 - r 1 n + 1 n + cos n n + 1 r n n + 1 + r 1 n + 1 . By the Squeeze Theorem, it suﬃces to show that lim n →∞ ³ r n n + 1 - r 1 n + 1 ! = 1 and lim n →∞ ³ r n n + 1 + r 1 n + 1 ! = 1 . 1

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So by linearity of limits, it suﬃces to show that lim n →∞ r n n + 1 = 1 and lim n →∞ r 1 n + 1 = 0. Notice that by the quotient and linearity theorems and the fact that n → ∞ implies 1 n 0 (Theorem 5 . 1 ), we have lim n →∞ n n + 1 = lim n →∞ ± n n + 1 · 1 n 1 n ² = lim n →∞ 1 1 + 1 n = lim n →∞ 1 lim n →∞ 1 + lim n
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## This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).

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hw4sol - Math 3110 Homework 4 Solutions Exercise 5.1.4 an...

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