This preview shows page 1. Sign up to view the full content.
Unformatted text preview: to 1 while s n does not converge. 2. Prove that if lim x n = x and if x > 0, then there exists a natural number M such that x n > for all n M . Set = x/ 2. This is positive by assumption. Applying the denition of convergence, there is M N so that  x nx  < x/ 2 , n M. orx/ 2 < x nx < x/ 2 . The left inequality implies that x n > x/ 2 > 0 for n M . 1...
View
Full
Document
This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

Click to edit the document details