This preview shows page 1. Sign up to view the full content.
Unformatted text preview: c . On the other hand, the subsequence consisting of the odd n has f ( x n ) = 0 0. Since c 6 = 0, we have that { f ( x n ) } does not converge as desired. 2. Let f,g be dened on A R to R , and let c be a cluster point of A . Suppose that f is bounded on a neighborhood of c and that lim x c g = 0. Prove that lim x c fg = 0. Fix > 0. Since f is bounded on a neighborhood of c , there exists a real number M and a 1 > 0 so that  f ( x )  M if  xc  < 1 and x A . Since lim x c g = 0, we have that there exists a 2 > 0 so that  g ( x )  < if 0 <  xc  < 2 and x A . Setting = min( 1 , 2 ), we see that  f ( x ) g ( x )  M  g ( x )  M if x A and 0 <  xc  < . Thus, f ( x ) g ( x ) 0. 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