1873_solutions

# Of the set s unequal to a and within a distance of a

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Unformatted text preview:  x 1 5x 2  4 3 5  7. Given that fx  1 if x is rational 0 if x is irrational explain why f does not have a limit from the right at 2. Solution: To obtain a contradiction, suppose that  is a limit of the function f from the right at 2. Using the fact that 1/2  0 we choose a number   0 such that the inequality |f x  |  1 2 will hold for every number x 2, 2   . Choose a rational number x and an irrational number t in the interval x, x   . Then 1  |f x f t | |f x  |  | f t |  1  1  1 2 2 and we have reached the desired contradiction. 8. Suppose that a is an interior point of a set S of real numbers and that f : S R. Suppose that f x 0 as x a and that f x 1 as x a . Prove that the function f does not have a limit at the number a. Since x must be a limit point of each of the sets Ý, a S and S a, Ý , the desired result follows at once from Theorem 8.3.2. Some Exercises on Continuity 1. Given that f x  x2 1 x 3 191 for every number x, prove that the function f is continuous at the number 2. 1 All we have to show is that f x as x 2. This fact can be deduced directly in the same way 7 that we did the earlier exercises on limits or, if you prefer, it can be deduced at once from Theorem 8.5.4. 2. Given that x sin fx  1 x if x 0 if x  0 0 prove that the function f is continuous at the number 0. Hint: Use the fact that |f x | |x | for every number x and use the sandwich theorem. The graph of this function is illustrated in the following figure. From the inequality 0 |f x f 0 | |x | that holds for every number x and from the sandwich theorem we see at once that f x x 0. f 0 as 3. Given that f is the ruler function defined earlier, explain why f is continuous at every irrational number in the interval 0, 1 and discontinuous at every rational number in 0, 1 . Solution: Since the equation f x  0 holds for x 0, 1 if and only if the number x is irrational, all we have to do is show that for every number x 0, 1 we have lim f t  0. tx Suppose that x 0, 1 and suppose that  0. Choose an integer N  1/ . Since there are only finitely many rational numbers that have a reduced form m/n for which n N we know that there are at most finitely many numbers x in the interval 0, 1 for which f x . We shall call this finite set S. Since the set S x , being finite, is closed, the set R S x must be open and is therefore a neighborhood of x. We also know that whenever t 0, 1 0 as t |f t x. R S x we have and so we have shown that f t 0|  f t  4. Suppose that f and g are functions from a given set S of real numbers into R and that the inequality |f t f x | |g t g x | holds for all numbers t and x in S. Prove that f must be continuous at every number at which the function g is continuous. Suppose that the function g is continuous at a given number x. Suppose that  0. Choose   0 such that the inequality |g t g x |  will hold whenever t S x , x   . Then for every such number t we have 192 |f t fx | gx | . |g t 5. Given that f is a continuous function...
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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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