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HW 6 Solutions

HW 6 Solutions - Math 117 F 09 HW#6(sample solutions...

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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 0 f ( x ); we want to prove that lim x x 0 | f ( x ) | = | L | . So let > 0. Since lim x x 0 f ( x ) = L , there exists δ > 0 such that | f ( x ) - L | < for all x D with 0 < | x - x 0 | < δ . By the identity ( ), we then get | f ( x ) | - | L | ≤ | f ( x ) - L | < for all x D with 0 < | x - x 0 | < δ . This proves that | L | = lim x x 0 | f ( x ) | = lim x x 0 | 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 half-open 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 0 R if and only if x 0 is not an integer. First suppose that x 0 is not an integer. Set n = [ x 0 ]; then x 0 ( n, n + 1). Set δ 1 = min { x 0 - n, n + 1 - x 0 } , so that δ 1 > 0 and ( x 0 - δ 1 , x 0 + δ 1 ) ( n, n + 1). In particular, f ( x ) = x - n for all x ( x 0 - δ 1 , x 0 + δ 1 ). We claim that lim x x 0 f ( x ) = x 0 - n . Now let > 0. Set δ = min { δ 1 , } . For all x R with 0 < | x - x 0 | < δ , we have x ( x 0 - δ 1 , x 0 + δ 1 ) and f ( x ) = x - n , and so | f ( x ) - ( x 0 - n ) | = | ( x - n ) - ( x 0 - n ) | = | x - x 0 | < δ . This proves that lim x x 0 f ( x ) = x 0 - n . Now suppose that x 0 is an integer, say x 0 = n . If f did have a limit at x 0 , say L , then there would exist a δ > 0 such that | f ( x ) - L | < 1 4 for all x R with 0 < | x - x 0 | < δ . Consider the numbers y = n - min δ 2 , 1 4 and z = n + min δ 2 , 1 4 .
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