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Unformatted text preview: Homework 3 NOTE: This was a halflength assignment, so only two problems were graded: 8.8 (worth 10 points) and 9.6 (worth 15 points). The other problems were graded for completion (worth 5 points each excpt for 9.1, which was worth 10) to bring the total score to 50 points. Scores were than doubled to make this assignment worth 100 points like the others. 8.8. Before beginning, we prove the following lemma. Lemma: Suppose A and B are strictly positive numbers and { c n } is a se quence of strictly positive numbers which converges to C . Then A B + c n converges to A B + C . Proof: We note that  A B + c n A B + C  =  A ( B + C ) A ( B + c n ) ( B + c n )( B + C )  ≤ A  C c n  B ( B + C ) ; it suffices to show that this last expression converges to zero. Fixing ǫ > we can pick N such that for n > N ,  C c n  < ǫ B ( B + C ) A (as c n → C ); this tells us that A  C c n  B ( B + C ) < ǫ A B ( B + C ) B ( B + C ) A = ǫ so our sequence is within ǫ of its desired limit; letting ǫ go to zero proves the lemma. Armed with this lemma, we now establish our limits. (a)  √ n 2 + 1 n  = 1  √ n 2 +1+ n  (multiplying numerator and denominator by √ n 2 + 1+ n ) ≤ 1 n + n (as √ is an increasing function and 1 /x < 1 /y if x > y > 0) = 1 2 n ; picking ǫ > 0 we therefore have that if n > 1 2 ǫ then  ( √ n 2 + 1 n )  < ǫ showing that the limit of...
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 Spring '09
 WEISBART
 9.3, strictly positive numbers

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