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Unformatted text preview: Math 117, F 09, HW#6 (sample solutions) P.79, #12. The symbol  f  denotes the function whose outputs are the absolute values of the outputs of f , that is,  f  : D → R and  f  ( x ) =  f ( x )  for all x ∈ D . We will need an identity about differences of absolute values. This is part (iv) of Theorem 0.5:  a    b  ≤  a b  ( † ) for any real numbers a and b . (If you have not seen this identity before, you should read the proof, or, better yet, make your own proof.) Now set L = lim x → x f ( x ); we want to prove that lim x → x  f ( x )  =  L  . So let > 0. Since lim x → x f ( x ) = L , there exists δ > 0 such that  f ( x ) L  < for all x ∈ D with 0 <  x x  < δ . By the identity ( † ), we then get  f ( x )    L  ≤  f ( x ) L  < for all x ∈ D with 0 <  x x  < δ . This proves that  L  = lim x → x  f ( x )  = lim x → x  f  ( x ). P.79, #13. The graph of this function is a “sawtooth”, which makes a jump at each integer. Note that for any integer n , the points x in the halfopen interval [ n,n +1) satisfy [ x ] = n . Thus, f ( x ) = x n for all x ∈ [ n,n + 1). Claim : f has a limit at a number x ∈ R if and only if x is not an integer. First suppose that x is not an integer. Set n = [ x ]; then x ∈ ( n, n + 1). Set δ 1 = min { x n, n + 1 x } , so that δ 1 > 0 and ( x δ 1 , x + δ 1 ) ⊆ ( n, n + 1). In particular, f ( x ) = x n for all x ∈ ( x δ 1 , x + δ 1 ). We claim that lim x → x f ( x ) = x n . Now let > 0. Set δ = min { δ 1 , } . For all x ∈ R with 0 <  x x  < δ , we have x ∈ ( x δ 1 , x + δ 1 ) and f ( x ) = x n , and so  f ( x ) ( x n )  =  ( x n ) ( x n )  =  x x  < δ ≤ . This proves that lim x → x f ( x ) = x n ....
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This note was uploaded on 08/27/2011 for the course MATH 117 taught by Professor Akhmedov,a during the Fall '08 term at UCSB.
 Fall '08
 Akhmedov,A
 Math

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