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ans5 - MAA 4200 Homework 5 Solutions[3.2 We must show that...

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MAA 4200: Homework 5 Solutions June 29, 2009 [3.2] We must show that lim x →- 3 f ( x ) = - 12. Preliminary Work: For x 6 = - 3, | f ( x ) + 12 | = | 2( x - 3) + 12 | = | 2 x + 6 | = 2 | x + 3 | . Proof: Let > 0. Choose δ = / 2. Then | f ( x ) + 12 | = 2 | x + 3 | < 2 δ = for all 0 < | x + 3 | < δ . By Theorem 3 . 1, f is continuous at x = - 3. [3.6] Prove that x : [0 , ) is continuous. Proof : First consider x 0 = 0. Let > 0. Choose δ = 2 . Then for all x D with | x | < δ we have, | x | = | x | 1 / 2 < δ 1 / 2 = . Now we consider x 0 > 0. Let > 0. Choose δ = x 0 . Then for all x D with | x - x 0 | < δ we have, | x - x 0 | = 1 x + x 0 | x - x 0 | ≤ 1 x 0 | x - x 0 | < 1 x 0 δ = . [3.9] As proved in problem 2 . 6 in previous homework, lim x 0 x sin(1 /x ) = 0, so defining f (0) = 0 will make f continuous at 0 by Theorem 3 . 1. [3.11] In problem 2 . 14 in previous homework, we proved that lim x 1 f ( x ) does not exist, hence f is discontinuous at x = 1 by Theorem 3 . 1. In problem 2 . 14 in previuos homework, we proved that lim x 2 f ( x ) = 16 and f (2) = 8 · 2 = 16 since 2 is rational. Therefore f is continuous at x = 2 by Theorem 3 . 1.

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