s090216 - Math 312, Intro. to Real Analysis: Homework #3...

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Unformatted text preview: Math 312, Intro. to Real Analysis: Homework #3 Solutions Stephen G. Simpson Monday, February 16, 2009 The assignment consists of Exercises 7.3, 7.5, 8.2(b)(e), 8.3, 8.8, 9.3, 9.4, 9.12, 9.18 in the Ross textbook. Each problem counts 10 points. 7.3. (a) 1, (b) 1, (c) 0, (d) 1, (e) does not converge, (f) 1, (g) diverges to + , (h) does not converge, (i) 0, (j) 7 / 2, (k) diverges to + , (l) does not converge, (m) 0, (n) does not converge, (o) 0, (p) 2, (q) 0, (r) 1, (s) 4 / 3, (t) 0. 7.5. (a) lim n 2 + 1 n = lim 1 n 2 + 1 + n = 0. (b) lim n 2 + n n = lim n n 2 + n + n = lim 1 radicalbig 1 + 1 /n + 1 = 1 2 . (c) lim radicalbig 4 n 2 + n 2 n = lim n 4 n 2 + n + 2 n = lim 1 radicalbig 4 + 1 /n + 2 = 1 4 . 8.2. (b) The limit is 7 / 3. Given > 0 we want vextendsingle vextendsingle vextendsingle vextendsingle 7 n 19 3 n + 7 7 3 vextendsingle vextendsingle vextendsingle vextendsingle < . (1) This is equivalent to vextendsingle vextendsingle vextendsingle vextendsingle 106 9 n + 21 vextendsingle vextendsingle vextendsingle vextendsingle < , i.e., 106 9 n + 21 < , i.e., n > 106 9 7 3 . The above reasoning works in reverse, so we have shown that (1) holds for all n > N where N = 106 / 9 7 / 3. This proves the limit....
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s090216 - Math 312, Intro. to Real Analysis: Homework #3...

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