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

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
This is the end of the preview. Sign up to access the rest of the document.

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 &gt; 0 we want vextendsingle vextendsingle vextendsingle vextendsingle 7 n 19 3 n + 7 7 3 vextendsingle vextendsingle vextendsingle vextendsingle &lt; . (1) This is equivalent to vextendsingle vextendsingle vextendsingle vextendsingle 106 9 n + 21 vextendsingle vextendsingle vextendsingle vextendsingle &lt; , i.e., 106 9 n + 21 &lt; , i.e., n &gt; 106 9 7 3 . The above reasoning works in reverse, so we have shown that (1) holds for all n &gt; N where N = 106 / 9 7 / 3. This proves the limit....
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online