1873_solutions

Sequence and then we shall observe that the limit of

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Unformatted text preview: t belong to Y. Therefore the metric space Y must be compact. 8 Limits and Continuity of Functions Some Exercises on Limits of Functions 183 1. Write careful proofs of each of the following assertions: a. x 3 3x 2 as x We observe 1. that if x is any given number we have | x 3 3x 2 |  |x 3 3x 2 |  | x  1 x2 x 2 | The idea of the proof is to make the observation that the factor x  1 will be small when x is close to 1 and that the factor x 2 x 2 is not too large. In fact, if |x  1|  1 then, since 2  x  0 we have |x 2 x 2 | 2 2  2  2  8 which gives us | x 3 3x 2 |  | x  1 x 2 x 2 | 8|x  1|. Now suppose that  0. We define to be the smaller of the two numbers 1 and /8. Then, whenever |x  1|   we have | x3 3x 2|  | x  1 x2 x 2| 8|x  1|  8 8 . We conclude that the inequality holds whenever x x  1). b. c. | x 3 3x 2 |  1 and |x  1|   (and, as a matter of fact, the inequality is also true when 1 as x 3. 3 We begin with the observation that if x is any nonzero number we have 1  |3 x | . 1 x 3 3|x | To keep the denominator of this fraction from becoming too small we need to keep x away from 0. In fact, if |x 3 |  1 then, since 2  x  4 we have 1  |3 x |  |3 x | . 1 x 3 6 3|x | Now suppose that  0. We define  to be the smaller of the two numbers 1 and and we observe that whenever |x 3 |   we have 1  |3 x |  |3 x |  1 . x 3 6 6 3|x | We conclude that the inequality 1 1 x 3 holds whenever x 3 and |x 3 |   (and, as a matter of fact, the inequality is also true when x  3). 1 x x3 8 12 as x 2. 5 x 6 We begin by observing that if x is any number for which x 2  x 6 2 x3 8 12  x 2 x  2x  4 5 x 2 x3 x2  x 6 x2 0 then we have 12 5 12 x 2  2x  4 5 x3 x 2 5x  8  5 x3 In order to keep the denominator of this fraction from becoming too small we need to keep x  184 away from 3 and, in order to keep |5x  8| from becoming too large we need to keep x from being too large. In fact, if x 2 and |x 2 |  1 we have x 2 5 3 8 x3 8 12  x 2 5x  8   |x 2 |. 5 5 x3 5 33 x2  x 6 Now suppose that  0 and define  to be the smaller of the two numbers 1 and . We see that whenever x 2 and |x 2 |   we have x3 8 12  |x 2 |  . 5 x2  x 6 2. Given that x prove that f x 1 as x if 0  x  2 x2 fx  if x  2 1 and that this function f has no limit at the number 2. Solution: 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 then, regardless of whether |f x 1 |  |x 1 | or |f x 1 |  |x 2 1 | we have |f x 1| 3|x 1|  3 3 With this fact in mind we define   /3 and we observe that |f x of f and x 1 and |x 1 |  . Note that we also have the inequality |f x 1|  2 and |x 1 |  /3 . 1|  whenever x lies in the domain 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...
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