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Solutions for the Exercises at the End of the Sections 3 255 x Since jx - 1j < , we have j x 2 - 1 - 3 j < j2x + 1j : We need to estimate the fraction j2x + 1j : The choice of 2jx + 1j 2jx + 1j 2 - 1 È 2 É = minimum 1, 5 implies that 1 and 2 : Thus jx - 1j < 1: So, -1 < x - 1 < 1, that is, 5 0 < x < 2: To get an estimate for the numerator, we notice that 0 < 2x < 4 and 1< 2x + 1 < 5: As an 5 estimate for the numerator, we have 1 < x + 1 < 3: Therefore, j2x + 1j < 2 × 1 = 5 : This allows us to 2jx + 1j 2 conclude that x 3 - 1 3 j2x + 1j 5 5 2 - < x 2 - 1 2 2jx + 1j < 2 2 × 5 = : È É This inequality implies that the choice of = minimum 1, 2 is indeed correct. 5 20. Let = 0:9: Then jx - 2j < = 0:9: This implies -0:9 < x - 2 < 0:9, that is, 1:1 < x < 2:9: Therefore, 3:3 < 3x < 8:7: Subtracting 6 yields -2:7 < ð3x - 5Þ -1 < 2:7: So, jð3x - 5Þ -1j < 2:7 < 4:5: Assume that jx - 2j < 1:5: Then -1:5 - < x -2 < 1:5: Adding 2 to all the parts of the inequality yields 0:5 2 - < x < 2 + 3:5: In turn, multiplying these inequalities by 3 implies that 1:5 3ð2 - Þ < 3x < 3ð2 + Þ 10:5: Subtracting 6 yields 1:5 - 6 < 3x - 6 < 10:5 - 6, i.e., -4.5 < 3x - 6 < 4.5. Therefore, jð3x - 5Þ - 1j < 4:5 whenever jx - 2j < 1:5: Sizes of Infinity 1. The set B is a proper subset of A if it has at most n - 1 of the elements of A. Let f : B A be a one- to-one function. Then f ðBÞ = Range f has exactly as many elements as B. Therefore, f ðBÞ = Range f