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ans4 - MAA 4200 Homework Solutions Sequential Limit Theorem...

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MAA 4200: Homework Solutions June 19, 2009 Sequential Limit Theorem = Theorem 2 . 1 Algebra of Limits Theorem = Theorem 2 . 4 [2.12] Prove that if lim x x 0 f ( x ) = L then lim x x 0 | f ( x ) | = | L | . Preliminary Work: | | f ( x ) | - | L | | ≤ | f ( x ) - L | for all x D by the triangle inequality. Proof: Let > 0. By definition of limit, there exists δ 1 > 0 such that | f ( x ) - L | < for 0 < | x - x 0 | < δ 1 . Choose δ = δ 1 and let 0 < | x - x 0 | < δ . Then | | f ( x ) | - | L | | ≤ | f ( x ) - L | < . [2.13] In Example 2.6 on p. 71 it is proved that lim x x 0 [ x ] exists iff x 0 is not an integer. Thus, since lim x x 0 exists for all x 0 R , the limit lim x x 0 ( x - [ x ]) exists for all x 0 / Z by the Algebra of Limits Theorem. We can also apply the Algebra of Limits Theorem to show that lim x x 0 ( x - [ x 0 ]) does not exist for x 0 Z . Let f ( x ) = x - [ x ]. Proof by contradiction: suppose lim x x 0 f ( x ) exists for some x 0 Z . Then lim x x 0 [ x ] = lim x x 0 ( x - f ( x )) exists by the Algebra of Limits Theorem, which is a contradiction. [2.14] Show that lim x 2 f ( x ) = 16 and the limit does not exist at any other point.
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