1873_solutions

F is a function from x into a metric space y that

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Unformatted text preview: /2  2 2 2 x2  y2 2 x2  y2 x2  y2 and, by considering the cases |x | and we conclude that f x, y |y | and |y | |x |, we see that |f x, y | |x |  |y | 0 as x, y 0, 0 . 11. Suppose that S is a set of real numbers, that a is a limit point of S, that f : S R k and that  that if f x  as x a then f x  as x a. Compare this exercise with the corresponding exercise for sequences. The key to this exercise is the fact that, whenever x is in the domain of f we have | fx  | fx and the proof runs along the same lines as the one used for sequences. 12. Suppose that X is a metric space, that S is a subset X and that both of the sets S and X Suppose that c X and that x if x S c if x fx  X R k . Prove S are dense in X. . S Prove that f has a limit at c but does not have a limit at any other point of the space X. To show that f x c as x c, suppose that  0. We define   and observe that whenever x c and d c, x   we have d c, f x  d c, x    when x S and we have d c, f x  d c, c  0  when x X S. Now suppose that x X c . We want to show that f has no limit at x. To obtain a contradiction, suppose that f has a limit  at x. We define  d c, x and note that  0. Choose   0 such that   /2, and such that for every point t X B c,  c , we have d f t ,   /4. Choose points t 1 and t 2 in the set X B x,  x such that t 1 S and t 2 X S. We see that d x, c d x, t 1  d t 1 , c  d t1, c 2 from which we conclude that d t 1 , c  /2. Therefore d f t1 , f t2  d t1, c   On the other hand, 215 2 . d f t1 , f t2 d f t 1 ,   d , f t 2  4  4  2 . This is the desired contradiction. 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. Given x 3, the inequality f x  w says that 1  w. |x 3 | We can’t simply turn these expressions over because w need not be positive but we can make the observation that the inequality 1 w |x 3 | will certainly hold when 1  |w |  1 |x 3 | which says that 1. |x 3 |  |w |  1 We therefore define   |w11 and observe that the condition f x  w will hold whenever x 3 and | |x 3 |  . fx  3. Given that 1 x3 for all numbers x 3, explain why f has an infinite limit from the left at 3 and also has an infinite limit from the right at 3 but does not have a limit at 3. The reason f has no two-sided limit at 3 is that the limits of f at 3 from the left and from the right are not equal to each other. In fact,...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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