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Unformatted text preview: ² > 0 there is N = N ( ² ) with ν ≤ s N < ν + ² . Since s n is nonincreasing, ν ≤ s n < ν + ² for all n > N . Thus  s nν  < ² for all n > N . Hence lim s n = ν by deﬁnition. In other words, { s n } converges to inf { s n } . 1 Problem A . (1) lim 3 n 2 + 6 2 n 2 + 5 n + 1 = lim 3 + 6 n 2 2 + 5 n + 1 n 2 = lim(3 + 6 n 2 ) lim(2 + 5 n + 1 n 2 ) = lim3 + lim 6 n 2 lim2 + lim 5 n + lim 1 n 2 = 3 2 . The same way (2) lim 3 n 3 + 6 2 n 2 + 5 n + 1 = lim 3 n + 6 n 2 2 + 5 n + 1 n 2 = ∞ 2 = ∞ . (3) lim 3 n 3 + 6 2 n 4 + 5 n + 1 = lim 3 n + 6 n 4 2 + 5 n 3 + 1 n 4 = 2 = 0. 2...
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This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Penn State.
 Spring '08
 OLSAVSKYGREGORM

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