Chapter 2 Solutions Manual

Chapter 2 Solutions Manual - CHAPTER 2 Limits and...

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27 CHAPTER 2 Limits and Continuity EXERCISE SET 2.1 1. (a) 0 (b) 0 (c) 0 (d) 3 3. (a) −∞ (b) −∞ (c) −∞ (d) 1 5. for all x 0 6 = 4 13. (a) 2 1 . 5 1 . 1 1 . 01 1 . 001 0 0 . 5 0 . 9 0 . 99 0 . 999 0 . 1429 0 . 2105 0 . 3021 0 . 3300 0 . 3330 1 . 0000 0 . 5714 0 . 3690 0 . 3367 0 . 3337 1 0 02 The limit is 1 / 3. (b) 2 1 . 5 1 . 1 1 . 01 1 . 001 1 . 0001 0 . 4286 1 . 0526 6 . 344 66 . 33 666 . 3 6666 . 3 50 0 12 The limit is + . (c) 0 0 . 5 0 . 9 0 . 99 0 . 999 0 . 9999 1 1 . 7143 7 . 0111 67 . 001 667 . 0 6667 . 0 0 –50 01 The limit is −∞ . 15. (a) 0 . 25 0 . 1 0 . 001 0 . 0001 0 . 0001 0 . 001 0 . 1 0 . 25 2 . 7266 2 . 9552 3 . 0000 3 . 0000 3 . 0000 3 . 0000 2 . 9552 2 . 7266

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28 Chapter 2 3 2 –0.25 0.25 The limit is 3. (b) 0 0 . 5 0 . 9 0 . 99 0 . 999 1 . 5 1 . 1 1 . 01 1 . 001 1 1 . 7552 6 . 2161 54 . 87 541 . 1 0 . 1415 4 . 536 53 . 19 539 . 5 60 –60 –1.5 0 The limit does not exist. 17. m sec = x 2 1 x +1 = x 1 which gets close to 2as x gets close to 1, thus y 1= 2( x +1)or y = 2 x 1 19. m sec = x 4 1 x 1 = x 3 + x 2 + x + 1 which gets close to 4 as x gets close to 1, thus y 1=4( x 1) or y =4 x 3 21. (a) The limit appears to be 3. 3.5 2.5 11 (b) The limit appears to be 3. 3.5 2.5 –0.001 0.001 (c) The limit does not exist. 3.5 2.5 –0.000001 0.000001 23. (a) The plot over the interval [ a, a ] becomes subject to catastrophic subtraction if a is small enough (the size depending on the machine). (c) It does not.
Exercise Set 2.3 29 25. (a) The length of the rod while at rest (b) The limit is zero. The length of the rod approaches zero EXERCISE SET 2.2 1. (a) 6 (b) 13 (c) 8 (d) 16 (e) 2 (f) 1 / 2 (g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. 3. 6 5. 3 / 4 7. 4 9. -4/5 11. 3 13. 3/2 15. + 17. does not exist 19. −∞ 21. + 23. does not exist 25. + 27. + 29. 6 31. (a) 2 (b) 2 (c) 2 33. (a) 3 (b) y x 4 1 35. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract inFnities. (b) lim x 0 + µ 1 x 1 x 2 = lim x 0 + µ x 1 x 2 = −∞ 37. lim x 0 x x ( x +4+2 ) = 1 4 39. The left and/or right limits could be plus or minus inFnity; or the limit could exist, or equal any preassigned real number. ±or example, let q ( x )= x x 0 and let p ( x a ( x x 0 ) n where n takes on the values 0 , 1 , 2. EXERCISE SET 2.3 1. (a) −∞ (b) + 3. (a) 0 (b) 1 5. (a) 12 (b) 21 (c) 15 (d) 25 (e) 2 3 / 5 (g) 0 (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. 7. −∞ 9. + 11. 3/2 13. 0 15. 0 17. 5 1 / 3 / 2 19. 5 21. 1 / 6 23. 3 25. −∞ 27. 1 / 7

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30 Chapter 2 29. It appears that lim t + n ( t )=+ , and lim t + e ( t )= c . 31. (a) + (b) 5 33. lim x + ( p x 2 +3 x ) x 2 +3+ x x 2 x = lim x + 3 x 2 x =0 35. lim x + ³ p x 2 + ax x ´ x 2 + ax + x x 2 + ax + x = lim x + ax x 2 + ax + x = a/ 2 37. lim x + p ( x )=( 1) n and lim x →−∞ p ( x 39. If m>n the limits are both zero. If m = n the limits are both equal to a m , the leading coeﬃcient of p .I f n>m the limits are ±∞ where the sign depends on the sign of a m and whether n is even or odd.
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Chapter 2 Solutions Manual - CHAPTER 2 Limits and...

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