1873_solutions

1 and x 1 and as a matter of fact the inequality

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Unformatted text preview: e that x fx  3 if x  3 x3 if x  3 To obtain a contradiction, suppose that the function f has a limit  at 3. Choose   0 such that   1 and such that whenever |x 3 |   and x 3 we have |f x  |  1. Choose numbers x 1 and x 2 such that 3   x 1  3  x 2  3  . We observe that f x 2 | |f x 1  |  |f x 2  |  1  1  2. |f x 1 On the other hand f x1  x1 3  2 3  5 and f x2  x2  3  6 and so f x2 |  6 5  2. |f x 1 This is the desired contradiction. 5. Given that S is a set of real numbers, that f : S R, that  is a real number and that a is a limit point of S, prove that the following conditions are equivalent: a. f x  as x a. b. For every number  0 there exists a number   0 such that the inequality |f x every number x in the set S a that satisfies the inequality |x a |  .  |  3 holds for c. For every number  0 there exists a number   0 such that the inequality |f x every number x in the set S a that satisfies the inequality |x a |  5.  |  3 holds for Solution: We shall provide the proof that condition c implies condition a. Suppose that condition c holds. To prove that condition a holds, suppose that  0. Using the fact that the number /3 is positive we now apply condition c to choose a number   0 such that the inequality |f x holds whenever x |x a |  . S a and |x |  3 3 a |  5. We see that |f x |  whenever x S a and 6. Given that S is a set of real numbers, that f : S R, that  is a real number and that a is a limit point of S, prove that the following conditions are equivalent: 187 a. f x  as x a. b. For every number  0 there exists a neighborhood U of the number a such that the inequality a. |f x  |  holds for every number x in the set U S c. For every neighborhood V of the number  there exists a number   0 such that the condition f x holds for every number x in the set S a that satisfies the inequality |x a |  . V To show that condition a implies condition b we assume that f x  as x a. Suppose that  0. From condition a and the fact that the interval  ,   is a neighborhood of  we deduce that there exists a neighborhood U of a such that the condition f x  ,   holds whenever xUS a. To show that condition b implies condition a we assume that condition b holds. Suppose that V is a neighborhood of . Choose a number  0 such that  ,   V. Now, using condition b, choose a neighborhood U of a such that the condition f x  ,   holds whenever xUS a . Then, whenever x U S a we have f x V. To show that condition a implies condition c we assume that condition a holds. Suppose that V is a neighborhood of  and, using condition a, choose a neighborhood U of a such that the condition fx V will hold whenever x U S a . Choose   0 such that a , a   U. We observe that whenever x S a and |x a |  , we must have x U S a and so f x V. To show that condition c implies condition a we assume that condition c holds. Suppose that V is a neighborhood of . Using condition c we choose a number   0 suc...
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