1873_solutions

# This limit is 0 note also that if x 0 then f x 0 0

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Unformatted text preview: the limit from the right is Ý and the limit from the left is Ý. To see why the f x Ý as x 3 , suppose that w is any real number. Given x  3, the inequality f x  w says that 1  w. x3 We can’t simply turn these expressions over because w need not be positive but we can make the observation that the inequality fx  216 1 x 3 w will certainly hold when 1 x  |w |  1 3 which says that 1. |w |  1 and observe that the condition f x  w will hold whenever 3 x We therefore define   |w11 | 3  x  3. To see why the f x Ý as x f x  w says that 3 , suppose that w is any real number. Given x  3, the inequality 1 x 3 w which we can express as 1 w 3x 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 3x will certainly hold when 1  |w |  1 x3 which says that 1. 3 x |w |  1 We therefore define   |w11 and observe that the condition f x  w will hold whenever | 3   x  3. 4. Prove that x3 8 x2  x 6 as x Ý. We begin by observing that x 2  x Ý 6  0 whenever x  2. Given any number x  2 we have 8  x 2  2x  4  x 2  x. x x3 x2  x 6 Now suppose that w is any real number and define v to be the larger of the two numbers 2 and w. The inequality x3 8  w 2x 6 x will hold whenver x  v. x3 5. Prove that x 4 4x 3 x 2  x  7 x 3 2x 2 2x 3 Ý as x Ý. Given any number x we have x3 2x 2 2x 3 x 3 x2  x  1 2x 2 2x 3 will be positive whenever x  3. Now given any number x  3 we have x 4 4x 3 x 2  x  7  x 4 4x 3 x 2  x 4 4x 3 x 3  x 5  x 5 . x 3 x3 x3 x 3 2x 2 2x 3 Now suppose that w is any real number. We define v to be the larger of the two numbers 3 and and so x 3 217 w 5 3 and observe that the inequality x 4 4x 3 x 2  x  7  w x 3 2x 2 2x 3 will hold whenever x  v. 6. Prove that 3x 2  x 1 5x 2  4 as x Ý. Given any number x we have 3x 2  x 1 5x 2  4 Whenever x  17 5 |5x 17| . 5 5x 2  4  we observe that 3x 2  x 1 5x 2  4 Now suppose that 3 5 3 5 |5x 5 5x 2  5x 2 5 5x 3 5  0. As long as x  17 5  17| 4 17 4  5x  1 . 5x 25x 2 , the inequality 3 x 1 5 5x 2  4 1 will hold whenever 51x  which says that x  5 . We define v to be the larger of the two numbers 17 1 and 5 and oberve that the inequality 5 3x 2 holds whenever x  v. 3x 2  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 fun...
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