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Unformatted text preview: 92.305 Homework 2 Solutions October 2, 2009 Exercise 2.2.1. Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit. (a) lim 1 (6 n 2 +1) = 0. Proof: Let ǫ > 0 be given. We take any N ∈ N strictly greater than radicalbig ( ǫ − 1 − 1) / 6. Then n ≥ N implies n 2 > ( ǫ − 1 − 1) / 6, hence 6 n 2 > ǫ − 1 − 1, and 6 n 2 + 1 > ǫ − 1 . Therefore, 1 / (6 n 2 + 1) < ǫ . Noting also that 1 / (6 n 2 + 1) > > − ǫ , we see that N works. squaresolid Proof: To keep the algebra simple, take N = radicalbig ( ǫ − 1 ) / 6. (Note that 1 6 n 2 +1 < 1 6 n 2 , so if can make 1 6 n 2 less than ǫ , 1 6 n 2 +1 will be less than ǫ as well.) (b) lim (3 n +1) (2 n +5) = 3 2 . Proof: Let ǫ > 0 be given. We take any N ∈ N strictly greater than (13 − 10 ǫ ) / 4 ǫ . Then n ≥ N implies n > (13 − 10 ǫ ) / 4 ǫ , hence 4 nǫ > 13 − 10 ǫ , and  4 n + 10  ǫ = (4 n + 10) ǫ = 4 nǫ + 10 ǫ > 13 = − 13  . Therefore, vextendsingle vextendsingle vextendsingle vextendsingle 3 n + 1 2 n + 5 − 3 2 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 6 n + 2 − 6 n − 15 4 n + 10 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle − 13 4 n + 10 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ, as desired. squaresolid (c) lim 2 √ n +3 = 0. Proof: Let ǫ > 0 be given. We take any N strictly greater than (2 /ǫ ) 2 − 3. Then n ≥ N implies n > (2 /ǫ ) 2 − 3, hence n + 3 > (2 /ǫ ) 2 and √ n + 3 > 2 /ǫ . This gives 2 / √ n + 3 < ǫ , and noting also that 2 / √ n + 3 > > − ǫ shows that N works. squaresolid Exercise 2.2.4. Argue that the sequence 1 , , 1 , , , 1 , , , , 1 , , , , , 1 , 5 zeros , 1 , . . . does not converge to zero. For what values of ǫ > 0 does there exist a response N ? For which values of ǫ > 0 is there no suitable response?...
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 Spring '10
 MetCalfe
 Calculus, Convergence, Limit of a sequence, 2k

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