1873_solutions

Reached the desired contradiction 3 given that fx

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Unformatted text preview: h that the condition f x V will hold whenever x S a , a   a . Since the interval a , a   is a neighborhood of a, condition a must hold. 7. Given that S is a set of real numbers, that f : S R, that  is a real number and that a is an interior 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 that satisfies the inequality |x a |  . |  will hold for The only way in which condition b differs from the ,  form of the assertion that f x  as x a is that it requires that |f x  |  for all numbers within a distance  of a and unequal to a. It does not merely assert that |f x  |  when x is a member of the set S unequal to a and within a distance  of a. It is obvious that condition b implies condition a. To show that condition a implies condition b we assume that f x  as x a. Suppose that  0. Choose a number  1  0 such that the condition |f x  |  will hold whenever x S a 1, a  1 a . Now, using the fact that a is an interior point of S, choose a number  2  0 such that a  2 , a   2 S. We define  to be the smaller of the two numbers  1 and  2 and we observe that the inequality |f x  |  will hold for every number x that satisfies the inequality |x a |  . 8. Suppose that S is a set of real numbers, that a is a limit point of S, that f : S R and that  is a real number. Prove that if f x  as x a then |f x | | | as x a. Compare this exercise with an earlier exercise. The key to this exercise is the fact that if x is any number in S then ||f x | | || |f x  |. To show that |f x | | | as x a, suppose that  0. Choose   0 such that the condition a , a   a . Then for all such numbers x we have |f x  |  will hold whenever x S ||f x | | || |f x  |  . 9. Suppose that S is a set of real numbers, that a is a limit point of S, that f : S R and that  is a real number. Complete the following sentence: The function f fails to have a limit of  at the number a when there exists a number  0 such that for every number   0 ...... The function f fails to have a limit of  at the number a when there exists a number  0 such that for every number   0 there is at least one number x in the set S a , a   a for which 188 |f x | . Some Further Exercises on Limits 1. Given that fx  1 if x  2 0 if x  2 prove that f has a limit from the left at 2 and also has a limit from the right at 2 but does not have a limit at 2. The fact that f does not have a limit at 2 will be clear when we have seen that f x 1 as x 2 and f x 0 as x 2 . Suppose that  0. We define   3 (or just take  to be any positive number you like). Whenever x  2 and |x 2 |   we have |f x 1 |  |1 1 |  0  and whenever x  2 and |x 2 |   we have |f x 0 |  |0 0 |  0  . 2. Given that 1 |x 3 | for all numbers x 3, explain why f has a limit (an infinite limit) at 3. We need to show that f x Ý as x 3. Suppose that w is a real number....
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