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Unformatted text preview: . So by the Squeeze Theorem, lim x f ( x ) = 0, and thus lim n a n = 0 as well. 2. (3 pts) b n = 7 n n ! There are many ways to show that this approaches 0. As one possibility, observe that whatever b 13 is, it is positive, and that for n 14 we have: b n b n1 = 7 n n ! ( n1)! 7 n1 = 7 n 1 2 So b 14 (1 / 2) b 13 , and b 15 (1 / 2) b 14 (1 / 2) 2 b 13 , and b 16 (1 / 2) b 15 (1 / 2) 3 b 13 , and so on. In general, for any positive n we have 0 b 13+ n (1 / 2) n b 13 , and we may apply the Squeeze Theorem to show that lim n b n = 0....
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This note was uploaded on 10/20/2010 for the course MATH 53903 taught by Professor Christ during the Spring '09 term at University of California, Berkeley.
 Spring '09
 CHRIST

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