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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 1 - 1.3 Number 25 – 51

Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 Section 1.1

CHAPTER 1 LIMITS

SECTION 1.3 Limit Theorems


Problem Set 1.3, Number 25 – 51.

Calculus 9th Purcell Chapter 1 - 1.3

    In Problems 25-30, find the limits if limx→a f(x) = 3 and limx→a g(x) = –1 (see Example 4).
  1. lim
    √[ f2(x) + g2(x)]
    x → a
     
    2. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  2. lim
    2f(x) – 3g(x)
    x → a
    f(x) + g(x)
    3. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  3. lim
    ∛(g(x)) ⋅ [f(x) + 3]
    x → a
     
    4. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  4. lim
    [f(x) – 3]⁴
    x → a
     
    5. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  5. lim
    [ |f(t)) + |3g(t)| ]
    t → a
     
    6. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  6. lim
    [f(u) + 3g(u)]3
    u → a
     
    7. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


    Semoga Bermanfaat 😁

    In Problems 31-34, find limx→2 [f(x) – f(2)] / (x – 2) for each given function f.
  7. f(x)
    =
    3x2
    8. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  8. f(x)
    =
    3x2 + 2x + 1
    9. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  9. f(x)
    =
    1
    x
    10. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  10. f(x)
    =
    3
    x2
    11. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  11. Prove Statement 6 of Theorem A. Hint:
    12. |f(x) g(x) – LM
    =
    |f(x) g(x) – Lg(x) + Lg(x) – LM
     
    =
    |g(x) [f(x) – L] + L [g(x) – M]|
     
    ≤
    |g(x)| |f(x) – L| + |L| |g(x) – M|
    Now show that if limx→c g(x) = M, then there is a number δ1 such that
    0 < |x – c| < δ1 ⇒ |g(x)| < |M| + 1
    ❤
    PEMBAHASAN:
    Suppose limx→c f(x) = L and limx→c g(x) = M.
    |f(x) g(x) – LM| ≤ |g(x)||f(x) – L| + |L||g(x) – M| as shown in the text.
    Choose ε1 = 1.
    Since limx→c g(x) = M, there is some δ1 > 0 such that if 0 < |x – c| < δ1, |g(x) – M| < ε1 = 1 or M – 1 < g(x) < M + 1.
    |M – 1| ≤ |M| + 1 and |M + 1| ≤ |M| + 1 so for 0 < |x – c| < δ1, |g(x)| < |M| + 1.
    Choose ε > 0.
    Since limx→c f(x) = L and limx→c g(x) = M, there exist δ2 and δ3 such that 0 < |x – c| < δ2 ⇒ |f(x) – L| < ε / (|L| + |M| + 1) and 0 < |x – c| < δ3 ⇒ |g(x) – M| < ε / (|L| + |M| + 1).
    Let δ = min {δ1, δ2, δ3}, then 0 < |x – c| < δ ⇒ |f(x) g(x) – LM| ≤ |g(x)||f(x) – L| + |L||g(x) – M| < (|M| + 1) ε / (|L| + |M| + 1) + |L| ε / (|L| + |M| + 1) = ε.
    Hence,
    limx→c f(x)g(x) = LM = (limx→c f(x))(limx→c g(x))

  12. Prove Statement 7 of Theorem A by first giving an ε – δ proof that limx→c [1/g(x)] = 1/[limx→c g(x)] and then applying Statement 6.
    13. ❤
    PEMBAHASAN:
    Say limx→c g(x) = M, M ≠ 0, and choose ε1 = ½|M|.
    There is some δ1 > 0 such that
    0 < |x – c| < δ1 ⇒ |g(x) – M| < ε = ½|M| or M – ½|M| < g(x) < M + ½|M|.
    |M – ½|M|| ≥ |½|M|| and |M + ½|M|| ≥ |½|M|| so |g(x)| > ½|M| and 1/|g(x)| < 2/|M|.
    Choose ε > 0.
    Since limx→c g(x) = M there is δ2 > 0 such that
    0 < |x – c| < δ2 ⇒ |g(x) – M| < ½M2.
    Let δ = min {δ1, δ2, then
    0 < |x – c| < δ ⇒ |1/g(x) – 1/M|
    = |M – g(x)/g(x)M|
    = 1/|M||g(x)| ⋅ |g(x) – M| < 2/M2 ⋅ |g(x) – M|
    = 2/M2 ⋅ ½M2ε
    = ε
    Thus, limx→c 1/g(x) = 1/M = 1/ limx→c g(x).
    Using statement 6 and the above result,
    limx→c f(x)/g(x) = limx→c f(x) ⋅ limx→c 1/g(x) = limx→c f(x) ⋅ 1/ limx→c g(x) = limx→c f(x)/limx→c g(x).

    Semoga Bermanfaat 😁

  13. Prove that limx→c f(x) = L ⇔ limx→c [f(x) – L] = 0.
    14. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  14. Proof that limx→c f(x) = 0 ⇔ limx→c |f(x)| = 0.
    15. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  15. Proof that limx→c |x| = |c|.
    16. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  16. Find examples to show that if
    17. (a)
    limx → c [f(x) + g(x)] exists, this does not imply that either limx → c f(x) or limx → c g(x) exists;
    (b)
    limx → c [f(x) ⋅ g(x)] exists, this does not imply that either limx → c f(x) or limx → c g(x) exists.
    ❤
    PEMBAHASAN:
    a.
    If f(x) = (x + 1)/(x – 2), g(x) = (x – 5)/(x – 2) and c = 2, then limx → c [f(x) + g(x)] exists, but neither limx → c f(x) nor limx → c g(x)] exists.
    b.
    If f(x) = 2/x, g(x) = x and c = 0, then limx → c [f(x) + g(x)] exists, but neither limx → c f(x) does not exists.

  17. lim
    √(3 + x)
    x → –3+
    x
    18. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  18. lim
    √(π3 + x3)
    x → –π+
    x
    19. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


    Semoga Bermanfaat 😁

  19. lim
    x – 3
    x → –3+
    √(x2 – 9)
    20. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  20. lim
    √(1 + x)
    x → 1–
    4 + 4x
    21. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  21. lim
    (x2 + 1) 〚x〛
    x → 2+
    (3x – 1)2
    22. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  22. lim
    (x – 〚x〛)
    x → 3–
     
    23. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  23. lim
    x
    x → 0–
    |x|
    24. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  24. Lim
    〚x2 + 2x〛
    x → 3+
     
    25. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  25. Suppose that f(x)g(x) = 1 for all x and limx→a g(x) = 0. Prove that limx→a f(x) does not exist.
    26. ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  26. Let R be the rectangle joining the midpoint of the sides of the quadrilateral Q having vertices (±x, 0) and (0, ±1). Calculate.
    27. lim
    perimeter of R
    x → 0+
    perimeter of Q
    ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


  27. Let y = √x and consider the point M, N, O, and P with coordinates (1, 0), (0, 1), (0, 0), and (x, y) on the graph of y = √x, respectively. Calculate
    28. (a)
    lim
    perimeter of ∆ NOP
    x → 0+
    perimeter of ∆ MOP
    ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


    (b)
    lim
    area of ∆ NOP
    x → 0+
    area of ∆ MOP
    ❤
    PEMBAHASAN:

    Calculus 9th Purcell Chapter 1 - 1.3


Semoga Bermanfaat 😁

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