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# Ans-12 - Answers Assignment#12 Exercises 3.1 1(a 1(c 1(e 1...

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Answers, Assignment #12 Exercises 3.1 1. (a) - 1 (b) - 2 (c) - 1 (d) - 5 (e) - 1 4 (f) 1 2 (g) 4 (h) - 2 2. (a) 27 (b) does not exist (c) 19 (d) 3 3. (a) False. You need 0 < | x - c | , i.e., you need x ± = c . (b) False. This is backwards; for every V you need a deleted neighborhood U ··· . (c) False. c must be an accumulation point of D . See Theorem 1. (d) True. Let ±> 0. Suppose lim x c f ( x )= L . There exists δ> 0 such that | f ( x ) - L | whenever | x - c | , x D. Therefore, | f ( c + h ) - L | whenever | ( c + h ) - c | = | h | , c + h D which implies lim h 0 f ( c + h L . Suppose lim h 0 f ( c + h L . There exists 0 such that | f ( c + h ) - L | whenever | h | + h Now, choose any x such that x D and | x - c | . Set h = x - c . Then | h | = c + h and | f ( x ) - L | = | f ( c + h ) - L | . Therefore, lim x c f ( x L . (e) True. Theorem 2. (f) True. For any positive integer n , lim x c x n = c n . The result follows from the limits in Theorem 3. (g) False. lim x c P ( x ) Q ( x ) does not exist if Q ( c )=0 ,P ( c ) ± =0 . 4. If | x - 3 | < 1, then | x 2 - 5 x +6 | = | x - 2 || x - 3 | < 2 | x - 3 | . 1

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Now, if | x - 3 | < 1 8 , then | x - 3 | is also less than 1 and | x 2 - 5 x +6 | < 1 4 .
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Ans-12 - Answers Assignment#12 Exercises 3.1 1(a 1(c 1(e 1...

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