05 005 x2 x2 x2 1 x2 1 2x 2 2 x 1 005

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Unformatted text preview: of p. If n > m the limits are ±∞ where the sign depends on the sign of am and whether n is even or odd. 64. (a) p(x) = q (x) = x (b) p(x) = x, q (x) = x2 (c) p(x) = x2 , q (x) = x (d) p(x) = x + 3, q (x) = x 65. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real number. For example, let q (x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0, 1, 2. Exercise Set 2.3 66. 50 If m > n the limit is zero. If m = n the limit is cm /dm . If n > m the limit is ±∞, where the sign depends on the signs of cn and dm . EXERCISE SET 2.3 |f (x) − f (4)| = |x2 − 16| < if |x − 4| < δ . We get f (x) = 16 + = 16.001 at x = 4.000124998, which corresponds to δ = 0.000124998; and f (x) = 16 − = 15.999 at x = 3.999874998, for which δ = 0.000125002. Use the smaller δ : thus |f (x) − 16| < provided |x − 4| < 0.000125 (to six decimals). (a) |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.1 if and only if |x| < 0.05 |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.01 if and only if |x| < 0.005 (c) |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.0012 if and only if |x| < 0.0006 (a) x1 = (1.95)2 = 3.8025, x2 = (2.05)2 = 4.2025 (b) δ = min ( |4 − 3.8025|, |4 − 4.2025| ) = 0.1975 (a) x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 . . . (b) 4. |f (x) − f (3)| = |(4x − 5) − 7| = 4|x − 3| < 0.1 if and only if |x − 3| < (0.1)/4 = 0.0025 (b) 3. |f (x) − f (0)| = |x + 2 − 2| = |x| < 0.1 if and only if |x| < 0.1 (c) 2. (a) (b) 1. δ = min( |1 − 0.909090|, |1 − 1.111111| ) = 0.0909090 . . . 5. |2x − 8| = 2|x − 4| < 0.1 if |x − 4| < 0.05, δ = 0.05 6. |x/2 + 1| = (1/2)|x − (−2)| < 0.1 if |x + 2| < 0.2, δ = 0.2 7. |7x + 5 − (−2)| = 7|x − (−1)| < 0.01 if |x + 1| < 8. |5x − 2 − 13| = 5|x − 3| < 0.01 if |x − 3| < 9. 10. 1 1 ,δ= 700 700 1 1 ,δ= 500 500 x2 − 4 x2 − 4 − 4x + 8 −4 = = |x − 2| < 0.05 if |x − 2| < 0.05, δ = 0.05 x−2 x−2 x2 − 1 x2 − 1 + 2x + 2 − (−2) = = |x + 1| < 0.05 if |x + 1| < 0.05, δ = 0.05 x+1 x+1 1 1 ,δ= 9000 9000 11. if δ < 1 then x2 − 16 = |x − 4||x + 4| < 9|x − 4| < 0.001 if |x − 4| < 12. √ √ 1 |x − 9| |x − 9| x+3 < |x − 9| < 0.001 if |x − 9| < 0.004, δ = 0.004 if δ < 1 then | x − 3| √ =√ <√ 4 x+3 | x + 3| 8+3 13. if δ ≤ 1 then 14. |x − 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05 16. |(4x − 5) − 7| = |4x − 12| = 4|x − 3| < 17. |2x − 7 − (−3)| = 2|x − 2| < |x − 5| |x − 5| 11 − = ≤ < 0.05 if |x − 5| < 1, δ = 1 x5 5|x| 20 15. |3x − 15| = 3|x − 5| < if |x − 3| < if |x − 2| < 1 2 ,δ= 1 2 1 4 ,δ= 1 4 if |x − 5| < 1 3 ,δ= 1 3 51 18. 19. Chapter 2 |2 − 3x − 5| = 3|x + 1| < x2 + x − 1 = |x| < x if |x + 1| < 1 3 ,δ= if |x| < , δ = 1 3 20. x2 − 9 − (−6) = |x + 3| < x+3 21. if δ < 1 then |2x2 − 2| = 2|x − 1||x + 1| < 6|x − 1| < 22. if δ < 1 then |x2 − 5 − 4| = |x − 3||x + 3| < 7|x − 3| < 23. 1 3|x − | 1 1 3 < 18|x − 1 | < −3 = if δ < then 6 x |x| 3 24. If δ < 25. √ √ √ x+2 x−4 1 =√ < |x − 4| < | x − 2| = ( x − 2) √ 2 x+2 x+2 26. √ √ |x − 6| 1 x+3+3 =√ ≤ |x − 6| < | x + 3 − 3| √ 3 x+3+3 x+3+3 27. |f (x) − 3| = |x + 2 − 3| = |x − 1| < 28. If δ < 1 then |(x2 + 3x − 1) − 9| = |(x − 2)(x + 5)| < 8|x − 2| < 1 7 , δ = min(1, 1 ) 7 1 1 1 if |x − | < , δ = min( 1 , 18 ) 6 3 18 if |x − 4| < 2 , δ = 2 if |x − 6| < 3 , δ = 3 if |x − 2| < 1 , δ = min (1, 1 ) 8 8 √ √ 1 < 0.1 if x > 10, N = 10 2 x 1 x − 1| = | | < 0.01 if x + 1 > 100, N = 99 (b) |f (x) − L| = | x+1 x+1 1 1 (c) |f (x) − L| = 3 < if |x| > 10, x < −10, N = −10 x 1000 (a) |f (x) − L| = |f (x) − L| = x 1 −1 = < 0.01 if |x + 1| > 100, −x − 1 > 100, x < −101, N = −101 x+1 x+1 (a) 1 < 0.1, x > 101/3 , N = 101/3 x3 1 < 0.01, x > 1001/3 , N = 1001/3 x3 1 < 0.001, x > 10, N = 10 x3 (a) x2 1 = 1 − , x1 = − 1 + x2 1 (b) N= 32. (a) x1 = −1/ 3 ; x2 = 1/ 33. 1 < 0.01 if |x| > 10, N = 10 x2 31. if |x − 3| < , δ = min(1, 1 ) 6 if 0 < |x − 1| < , δ = (c) 30. 1 6 5 3 1 1 1 and |x − (−2)| < δ then − < x < − , x + 1 < − , |x + 1| > ; then 2 2 2 2 2 |x + 2| 1 1 − (−1) = < 2|x + 2| < if |x + 2| < , δ = min( 1 , 1 ) 22 x+1 |x + 1| 2 (d) 29. if |x − 1| < if |x + 3| < , δ = 1− ; (b) x2 2 = 1 − , x2 = 1 + x2 2 1− (c) N = − 3 (b) N = 1/ 3 1− 1− (c) N = −1/ 3 Exercise Set 2.3 34. 52 1 < 0.005 if |x + 2| > 200, x > 198, N = 198 x+2 35. x 1 −1 = < 0.001 if |x + 1| > 1000, x > 999, N = 999 x+1 x+1 36. 4x − 1 11 −2 = < 0.1 if |2x + 5| > 110, 2x > 105, N = 52.5 2x + 5 2x + 5 37. 1 − 0 < 0.005 if |x +...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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