analhwans6

analhwans6 - 2.2.1. Verify, using the denition of...

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2.2.1. following sequences converge to the proposed limit. 1. lim 1 6 n 2 + 1 = 0 . Given a positive number " > 0 N so that for each n N we have 1 6 n 2 + 1 < " . This inequality is 6 n 2 + 1 > 1 =" , which is certainly true if n 2 1 =" , that is, if n p 1 =" . So we can just choose any natural number N p 1 =" (by the archimedean property!). 2. lim 3 n + 1 2 n + 5 = 3 2 . Given a positive number " > 0 N so that for each n N we have 3 n + 1 2 n + 5 ± 3 2 < " . Rewriting di/erence of the two fractions on the left 3 n + 1 2 n + 5 ± 3 2 = 6 n + 2 ± 6 n ± 15 4 n + 10 = ± 13 4 n + 10 we see that we want 13 4 n + 10 < " 13 " < 4 n + 10 13 " ± 10 < 4 n 13 4 " ± 5 2 < n so we can pick N to be any natural number greater than 13 4 " ± 5 2 . 3. lim 2 p n + 3 = 0 . Given a positive number " > 0 natural number N so that for each n N we have 2 p n + 3 < " . We 1
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can rewrite this as 2 " < p n + 3 4 " 2 < n + 3 4 " 2 3 < n so we can pick N to be any natural number greater than
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This note was uploaded on 09/08/2011 for the course MAT 472 taught by Professor Spielberg during the Spring '06 term at ASU.

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analhwans6 - 2.2.1. Verify, using the denition of...

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