HW 7 Solutions

HW 7 Solutions - Math 117, F 09, HW#7 (sample solutions)...

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Math 117, F 09, HW#7 (sample solutions) P.80, #27. Set X = D [ x 0 - ±, x 0 + ± ] = E [ x 0 - ±, x 0 + ± ]. (These two sets are equal by hypothesis.) We are also assuming that f ( x ) = g ( x ) for all x X . Suppose that f has a limit at x 0 , say L = lim x x 0 f ( x ). We claim that then L is also a limit for g at x 0 . So, let ± 0 > 0. Since L is the limit of f at x 0 , there exists δ > 0 such that | f ( x ) - L | < ± 0 for all x D with 0 < | x - x 0 | < δ . Set δ 0 = min { δ,± } , which is also positive. If x E with 0 < | x - x 0 | < δ 0 , then because δ 0 ± , we have x E ( x 0 - ±, x 0 + ± ) X = D [ x 0 - ±, x 0 + ± ] D. Consequently, f ( x ) = g ( x ) and x D with 0 < | x - x 0 | < δ , so | f ( x ) - L | < ± 0 and thus | g ( x ) - L | < ± 0 . This proves that g has limit L at x 0 , as claimed. This proves one of the claimed implications: If f has a limit at x 0 , then g has a limit at x 0 . By switching the roles of f and g , we have the other implication: If g has a limit at x 0 , then f has a limit at x 0 . ± P.104, #1.
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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.

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HW 7 Solutions - Math 117, F 09, HW#7 (sample solutions)...

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