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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 n-x | < x/ 2 , n M. or-x/ 2 < x n-x < x/ 2 . The left inequality implies that x n > x/ 2 > 0 for n M . 1...
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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