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Unformatted text preview: Section 5.4 #2)  1 /x 2 1 /y 2  =  y x  x + y  / ( x 2 y 2 )  y x  1  x  y 2 + 1 x 2  y  . If x,y [1 , then both 1 xy 2 and 1 x 2 y are less than or equal to 1. Hence,  1 /x 2 1 /y 2  2  x y  . So f is Lipschitz on [1 , ), and hence uniformly continuous. Suppose f were uniformly continuous on (0 , ). Then in particular, it would be uniformly continuous on (0 , 1). By the Continuous Extension Property, it could be extended to a continuous function on [0 , 1]. But continuous functions on closed bounded intervals are bounded, which f is not. Contradiction. #4)  f ( x ) f ( y )  =  x y   x + y  (1+ x 2 )(1+ y 2 )  x y   y  (1+ x 2 )(1+ y 2 ) +  x  (1+ x 2 )(1+ y 2 ) . If  x  1, then  x  / ((1 + x 2 )(1 + y 2 )) 1. If  x  > 1, then  x  / ((1 + x 2 )(1 + y 2 )) <  x  / ( x 2 (1 + y 2 )) < 1 /  x  < 1. Hence,  y  (1+ x 2 )(1+ y 2 ) +  x  (1+ x 2 )(1+ y 2 ) 1 + 1 = 2. So  f ( x ) f ( y )  2  x y  . So f is Lipschitz on R , and hence uniformly continuous. #7) It is obvious that f ( x ) = x is uniformly continuous. For all x,y R ,  sin( x ) sin( y )  = 2  sin( x y 2 )  cos( x + y 2 )...
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 JUNGE

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