1873_solutions

# Now since t n x n 0 as n we deduce from the

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Unformatted text preview: Now we want to show that the function f has no limit at the number 2. To obtain a contradiction, suppose that  is a limit of the function f at 2. The key to the desired contradiction is the fact that when x  2 we have f x  x 2 which is close to 4 when x is close to 2 and when x  2 we have f x  x which is close to 2 when x is close to 2. We shall use this observation to argue that both of the numbers 2 and 4 must lie close to the limit value , in spite of the fact that the distance from 2 to 4 is 2. Using the fact that 1  0 and that  is a limit of f at 2, we choose a number   0 such that the inequality |f x  |  1 holds whenever x 2 and |x 2 |  . Choose a number x 1  2 and a number x 2  2 such that |x 1 2 |   and |x 2 2 |  . Then f x 2 |  |f x 1    f x2 | |f x 1  |  | f x 2 |  1  1  2. |f x 1 2 On the other hand, f x 1  x 1  2 and f x 2  x 2  4 which gives us f x 2 |  2. |f x 1 We have therefore reached the desired contradiction. 3. Given that fx  prove that f x 1 as x x if x is rational x 2 if x is irrational 1 and prove that this function f has no limit at the number 2. Solution: This solution is very similar to the solution of Exercise 2 but we have to exercise a little more care with the inequalities at the end. Before we prove that f x 1 as x 1 we make the observation that whenever |x 1 |  1 we have |x 2 1 |  |x 1 ||x  1|  3|x 1 |. To prove that f x 1 as x 1, suppose that  0. We observe that whenever x 2 and |x 1 |  /3 then, regardless of whether |f x 1 |  |x 1 | or |f x 1 |  |x 2 1 | we have . 3 With this fact in mind we define   /3 and we observe that |f x 1 |  domain of f and x 1 and |x 1 |  . |f x Note that we also have the inequality |f x 1| 3|x 1|  1|  3 whenever x lies in the when x  1 but we do not need this fact. Now we want to show that the function f has no limit at the number 2. To obtain a contradiction, suppose that  is a limit of the function f at 2. The key to the desired contradiction is the fact that when x is irrational we have f x  x 2 which is close to 4 when x is close to 2, and when x is rational we have f x  x which is close to 2 when x is close to 2. We shall use this observation to argue 207 that both of the numbers 2 and 4 must lie close to the limit value , in spite of the fact that the distance from 2 to 4 is 2. Using the fact that 1  0 and that  is a limit of f at 2, we choose a number   0 such that the inequalities |f x  |  1 2 1 2 4|  |x 2 1 |x 2 |  2 all hold whenever x 2 and |x 2 |  . Choose a rational number x 1 and an irrational number x 2 such that |x 1 2 |   and |x 2 2 |  . Then f x 2 |  |f x 1    f x2 | |f x 1  |  | f x 2 |  1  1  1. |f x 1 2 2 On the other hand, f x 1  x 1 and f x 2  x 2 2 and so f x 2 |  |f x 2 2|  1  1  1  3 4 2 |4 f x 2 |  |f x 1 2 2 2 2 f x 2 |  2. |f x 1 and we have therefore reached the desired contradiction. 4. Given that for every number x 2 9 fx  x |x 3 | 3, prove that f has no limit at the...
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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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