This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Section 3.1 #4) Let > 0 be arbitrary. Select K N such that K >  b  . Then if n K , we have b n =  b  n  b  K < . Therefore, lim n b n = 0. #6) (a) Let > 0 be arbitrary. Select K N large enough so that K > 1 / 2 . Then if n K , we have n + 7 > n K > 1 / 2 . It follows that if n K , then  1 / n + 7  < . Therefore, lim n 1 / n + 7 = 0. (b) 2 n n +2 2 = 4 n +2 = 4 n +2 < 4 n . Since 1 /n 0, we can apply Theorem 3.1.10 (with C = 4) to get the desired result. (c) n n +1 = n n +1 < n +1 n +1 = 1 n +1 . The limit of 1 n +1 is 0, (proved just like in part (a) of this problem), and so we can apply Theorem 3.1.10 (with C = 1) to get the desired result. (d) ( 1) n n n 2 +1 = n n 2 +1 < n n 2 = 1 n . Since 1 /n 0, we can apply Theorem 3.1.10 (with C = 1) to get the desired result. #8) lim n x n = 0 ( > 0)( K N )( n K )  x n  < ( > 0)( K N )( n K )  x n  < lim n  x n  = 0 . Define x n := ( 1) n . Then  x n  = 1 for all n , so lim  x n  = 1 even though ( x n ) has no limit. #13) < b < 1, so 1 /b > 1. Let a = (1 /b ) 1 > 0 so that b = 1 1+ a . (1 + a ) n > 1 2 n ( n 1) a 2 as in Example 3.1.11(d). So  nb n  = n (1+ a ) n < 2 ( n 1) a 2 . We can now apply Theorem 3.1.10 since 1 n 1 0 (we take C = 2 a 2 ). #15) X := ( x n = n 2 n ! : n N } . We prove that X 2 converges to 0. By Theorem 3.1.9, this shows that X converges to 0. For n 2, x n +1 x n = ( n +1) 2 ( n +1)!...
View
Full
Document
This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JUNGE
 Squeeze Theorem

Click to edit the document details