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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 1 - 1.2 Number 1 – 33

Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 Section 1.1

CHAPTER 1 LIMITS

SECTION 1.2 Rigorous Study of Limits


Problem Set 1.2, Number 1 – 33.

Calculus 9th Purcell Chapter 1 - 1.2

    In Problems 1 – 6, give the appropriate ε – δ definition, of each statement.
  1. limt → a f(t) = M
    2. ❤
    PEMBAHASAN:
    0 < |t – a| < δ ⇒ |f(t) – M| < ε

  2. limu → b g(u) = L
    3. ❤
    PEMBAHASAN:
    0 < |u – b| < δ ⇒ |g(u) – L| < ε

  3. limz → d h(z) = P
    4. ❤
    PEMBAHASAN:
    0 < |z – d| < δ ⇒ |h(z) – P| < ε

  4. limy → e Φ(y) = B
    5. ❤
    PEMBAHASAN:
    0 < |y – e| < δ ⇒ |Φ(y) – B| < ε

  5. limx → c– f(x) = L
    6. ❤
    PEMBAHASAN:
    0 < c – x < δ ⇒ |f(x) – L| < ε

  6. limt → a+ g(t) = D
    7. ❤
    PEMBAHASAN:
    0 < t – a < δ ⇒ |g(t) – D| < ε

    In Problem 7 – 10, plot the function f(x) over the interval [1.5, 2.5]. Zoom in on the graph of each function to determine how close x must be to 2 in order that f(x) is within 0.002 of 4. Your answer should be of the form “If x is within ______ of 2, then f(x) is within 0.002 of 4.”
  7. f(x) = 2x
    8. ❤
    PEMBAHASAN:
    If x is within 0.001 of 2, then 2x is within 0.002 of 4.

    Calculus 9th Purcell Chapter 1 - 1.2


  8. f(x) = x2
    9. ❤
    PEMBAHASAN:
    If x is within 0.0005 of 2, then x2 is within 0.002 of 4.

    Calculus 9th Purcell Chapter 1 - 1.2


  9. f(x) = √(8x)
    10. ❤
    PEMBAHASAN:
    If x is within 0.0019 of 2, then √8x is within 0.002 of 4.

    Calculus 9th Purcell Chapter 1 - 1.2


  10. f(x) = 8/x
    11. ❤
    PEMBAHASAN:
    If x is within 0.001 of 2, then 8/x is within 0.002 of 4.

    Calculus 9th Purcell Chapter 1 - 1.2


    Semoga Bermanfaat 😁

    In Problems 11 – 22, give an ε – δ, proof of each limit fact.
  11. Lim
    (2x – 1) = –1
    x → 0
     
    12. ❤
    PEMBAHASAN:
    0 < | x – 0| < δ
    ⇒
    |(2x – 1) – (–1)| < ε
    |2x – 1 + 1| < ε
    ⇔
    |2x| < ε
     
    ⇔
    2 |x| < ε
     
    ⇔
    |x| < ε/2
     
     
     
                    δ = ε/2; 0 < |x – 0| < δ
                    |(2x – 1) – (–1)| = |2x| = 2 |x| < 2δ = ε

  12. Lim
    (3x – 1) = –64
    x → -21
     
    13. ❤
    PEMBAHASAN:
    0 < | x + 21| < δ
    ⇒
    |(3x – 1) – (–64)| < ε
    |3x – 1 + 64| < ε
    ⇔
    |3x + 63| < ε
     
    ⇔
    |3(x + 21)| < ε
     
    ⇔
    3 |x + 21| < ε
     
    ⇔
    |x + 21| < ε/3
     
     
     
                    δ = ε/3; 0 < |x + 21| < δ
                    |(3x – 1) – (–64)| = |3x + 63| = 3 |x + 21| < 3δ = ε

  13. lim
    x → 5
    x2 – 25
    = 10
    x – 5
    14. ❤
    PEMBAHASAN:
    0 < | x + 21| < δ
    ⇒
     
    x2 – 25
    – 10
    < ε
     
    x – 5
     
     
    x2 – 25
    – 10
    < ε
    ⇔
     
    (x – 5) (x + 5)
    – 10
    < ε
     
     
    x – 5
     
    x – 5
     
     
     
     
    ⇔
    |(x + 5) – 10| < ε
     
     
     
     
    ⇔
    |x – 5| < ε
     
     
     
     
     
     
                    δ = ε; 0 < |x – 5| < δ
     
     
    x2 – 25
    – 10
    =
     
    (x – 5)(x + 5)
    – 10
    =
    |x + 5 – 10|
     
    x – 5
    x – 5
                = |x – 5| < δ = ε

  14. lim
    x → 0
    (
    2x2 – x
    )
    = – 1
    x
    15. ❤
    PEMBAHASAN:
    0 < | x – 0| < δ
    ⇒
     
    2x2 – x
    – (–1)
    < ε
     
    x
     
     
    2x2 – x
    + 1
    < ε
    ⇔
     
    x(2x – 1)
    + 1
    < ε
     
     
    x
     
    x
     
     
     
     
    ⇔
    |2x – 1 + 1| < ε
     
     
     
     
    ⇔
    |2x| < ε
     
     
     
     
    ⇔
    2 |x| < ε
     
     
     
     
    ⇔
    |x| < ε/2
     
     
     
     
     
     
                    δ = ε/2; 0 < |x – 0| < δ
     
     
    2x2 – x
    – (–1)
    =
     
    x(2x – 1)
    + 1
    =
    |2x – 1 + 1|
     
    x
    x
                = |2x| = 2|x| < 2δ = ε

  15. lim
    x → 5
    2x2 – 11x + 5
    = 9
    x – 5
    16. ❤
    PEMBAHASAN:
    0 < |x – 5| < δ
    ⇒
     
    2x2 – 11x + 5
    – 9
    < ε
     
    x – 5
     
     
    2x2 – 11x + 5
    – 9
    < ε
    ⇔
     
    (2x – 1)(x – 5)
    – 9
    < ε
     
     
    x – 5
     
    x – 5
     
     
     
     
    ⇔
    |2x – 1 – 9| < ε
     
     
     
     
    ⇔
    |2(x – 5)| < ε
     
     
     
     
    ⇔
    |x – 5| < ε/2
     
     
     
     
     
     
                    δ = ε/2; 0 < |x – 5| < δ
     
     
    2x2 – 11x + 5
    – 9
    =
     
    (2x – 1)(x – 5)
    – 9
     
    x – 5
    x – 5
                    = |2x – 1 – 9| = |2(x – 5)| = 2|x – 5| < 2δ = ε

  16. lim
    √(2x) = √2
    x → 1
     
    17. ❤
    PEMBAHASAN:
    0 < | x – 1| < δ
    ⇒
    |√2x – √2| < ε
    |√2x – √2| < ε
     
     
     
    ⇔
     
    (√2x – √2)(√2x + √2)
     
    < ε
     
     
    √2x + √2
     
    ⇔
     
    2x – 2
     
    < ε
     
     
    √2x + √2
     
    ⇔
    2
     
    x – 1
     
    < ε
     
     
    √2x + √2
                    δ = (√2)ε/2; 0 < |x – 1| < δ
    |√2x – √2|
    ⇒
     
    (√2x – √2)( √2x + √2)
     
     
    √2x + √2
    =
     
    2x – 2
     
     
    √2x + √2
    2 |x – 1|
    ≤
    2 |x – 1|
    < 
    2δ
    =
    ε
    √2x + √2
    √2
    √2

  17. Lim
    x → 4
    √(2x – 1)
    = √7
    √(x – 3)
    18. ❤
    PEMBAHASAN:
    0 < |x – 4| < δ
    ⇒
     
    √(2x – 1)
    – √7
    < ε
     
    √(x – 3)
     
     
    √(2x – 1)
    – √7
    < ε
    ⇔
     
    √(2x – 1) – √7(x – 3)
     
    < ε
     
     
    √(x – 3)
     
    √(x – 3)
    ⇔
     
    [√(2x – 1) – √(7(x – 3))]  [√(2x – 1) + √(7(x – 3))]
     
    < ε
     
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    ⇔
     
    2x – 1 – (7x – 21)
     
    < ε
     
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    ⇔
     
    –5 (x – 4)
     
    < ε
     
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    ⇔
    |x – 4| ·
    5
    < ε
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    To bound
    5
    , agree that
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    δ ≤
    1
    . If δ ≤
    1
    , then
    7
    < x <
    9
    , so
    2
    2
    2
    2
    0.65 <
    5
    < 1.65 and
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    hence
    |x – 4| ·
    5
    < ε
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    ⇔
    |x – 4| <
    ε
    1.65
    For whatever ε is chosen, let δ be the smaller of
    1
    and
    ε
    .
    2
    1.65
    δ = min
    {
    1
    ,
    ε
    }
    , 0 < |x – 4| < δ
    2
    1.65
     
    √(2x – 1)
    – √7
     
    = |x – 4| ·
    5
     
    √(x – 3)
     
    √(x – 3) [√(2x – 1) + √(7(x – 3))]
    < |x – 4|(1.65) < 1.65 δ ≤ ε
    Since δ =
    1
    only when
    1
    ≤
    ε
    so 1.65 δ ≤ ε.
    2
    2
    1.65

  18. lim
    x → 1
    14x2 – 20x + 6
    = 8
    x – 1
    19. ❤
    PEMBAHASAN:
    0 < |x – 1| < δ
    ⇒
     
    14x2 – 20x + 6
    – 8
    < ε
     
    x – 1
     
     
    14x2 – 20x + 6
    – 8
    < ε
    ⇔
     
    2(7x – 3) (x – 1)
    – 8
    < ε
     
     
    x – 1
     
    x – 1
     
     
    ⇔
    |2(7x – 3) – 8| < ε
     
     
    ⇔
    |14(x – 1)| < ε
     
     
    ⇔
    14|x – 1| < ε
     
     
    ⇔
    |x – 1| < ε/14
     
    δ = ε/14; 0 < |x – 1| < δ
     
     
    14x2 – 20x + 6
    – 8
    =
     
    2(7x – 3) (x – 1)
    – 8
     
     
    x – 1
     
    x – 1
    =
    |2(7x – 3) – 8|
    =
    |14(x – 1)| = 14|x – 1| 14δ = ε

  19. lim
    x → 1
    10x3 – 26x2 + 22x – 6
    = 4
    (x – 1)2
    20. ❤
    PEMBAHASAN:
    0 < |x – 1| < δ
    ⇒
     
    10x3 – 26x2 + 22x – 6
    – 4
    < ε
     
    (x – 1)2
     
     
    10x3 – 26x2 + 22x – 6
    – 4
    < ε
    ⇔
     
    (10x – 6) (x – 1)2
    – 4
    < ε
     
     
    (x – 1)2
     
    (x – 1)2
     
     
    ⇔
    |10x – 6 – 4| < ε
     
     
    ⇔
    |10(x – 1)| < ε
     
     
    ⇔
    10|x – 1| < ε
     
     
    ⇔
    |x – 1| < ε/10
     
    δ = ε/10; 0 < |x – 1| < δ
     
     
    10x3 – 26x2 + 22x – 6
    – 4
    =
     
    (10x – 6) (x – 1)2
    – 4
     
     
    (x – 1)2
     
    (x – 1)2
     
    =
    |10x – 6 – 4|
     
    =
    |10(x – 1)|
     
    =
    10|x – 1| < 10δ = ε

  20. lim
    (2x2 + 1) = 3
    x → 1
     
    21. ❤
    PEMBAHASAN:
    0 < |x – 1| < δ
    ⇒
    |(2x2 + 1) – 3| < ε
    |2x2 + 1 – 3| < ε
    =
    |2x2 – 2|
     
    =
    2 |x + 1| |x – 1|
    To bound |2x + 2|, agree that δ ≤ 1.
    |x – 1| < δ implies
    |2x + 2|
    =
    |2x – 2 + 4|
     
    ≤
    |2x – 2| + |4|
     
    < 
    2 + 4 = 6
    δ ≤ ε/6; δ = min{1, ε/6}; 0 < |x – 1| < δ
    |(2x2 + 1) – 3|
    =
    |2x2 – 2|
    = |2x + 2||x – 1| < 6 ‧ (ε/6) = ε

  21. lim
    (x2 – 2x – 1) = 2
    x → -1
     
    22. ❤
    PEMBAHASAN:
    0 < |x + 1| < δ
    ⇒
    |(x2 – 2x – 1| < ε
    |x2 – 2x – 1 – 2|
    =
    |x2 – 2x – 3|
     
    =
    |x + 1| |x – 3|
    To bound |x – 3|, agree that δ ≤ 1.
    |x + 1| < δ implies
    |x – 2|
    =
    |x + 1 – 4|
     
    ≤
    |x + 1| + |–4|
     
    < 
    1 + 4 = 5
    δ ≤ ε/5; δ = min{1, ε/5}; 0 < |x + 1| < δ
    |(x2 – 2x – 1) – 2|
    =
    |x2 – 2x – 3|
    = |x + 1||x – 3| < 5 ‧ (ε/5) = ε

    Semoga Bermanfaat 😁

  22. lim
    x4 = 0
    x → 0
     
    23. ❤
    PEMBAHASAN:
    0 < |x| < δ
    ⇒
    |x4 – 0| = |x4| < ε
    |x4|
    =
    |x||x3|
    To bound |x3|, agree that δ ≤ 1.
    |x| < δ ≤ 1 implies
    |x3|
    =
    |x|3 ≤ 1
     so δ
    ≤
    ε.
    δ = min{1, ε}; 0 < |x| < δ
    |x4|
    =
    |x||x3| < ε ‧ 1 = ε

  23. Prove that if limx→c f(x) = L and limx→c f(x) = M, then L = M.
    24. ❤
    PEMBAHASAN:
    Choose ε > 0. Then since limx→c f(x) = L, there is some δ1 > 0 such that
     
    0 < |x – c| < δ1
    ⇒
    |f(x) – L| < ε.
     
    Since limx→c f(x) = M, there is some δ2 > 0 such that
     
    0 < |x – c| < δ2
    ⇒
    |f(x) – M| < ε.
     
    Let δ = min{δ1, δ2} and choose x0 such that
     
    0 < |x0
    –
    c| < δ.
     
    Thus, |f(x0) – L| < ε
    ⇒
    –ε < f(x0) – L < ε
     
     
    ⇒
    – f(x0) – ε < L < – f(x0) + ε
     
    ⇒
    f(x0) – ε < L < f(x0) + ε.
    Similarly,
     
     
    f(x0) – ε <
    M
    < f(x0) + ε.
    Thus,
     
     
    –2ε < L – M < 2ε. As ε ⇒ 0, L – M → 0, so L = M.

  24. Let F and G be functions such that 0 ≤ F(x) ≤ G(x) for all x near c, except possibly at c. Prove that if limx→c f(x) = 0, then limx→c f(x) = 0.
    25. ❤
    PEMBAHASAN:
    Since limx→c G(x) = 0, then given any ε > 0, we can find δ1 > 0 such that whenever
    |x – c| < δ, |G(x)| < ε
    Take any ε > 0 and the corresponding δ that works for G(x), then |x – c| < δ implies
    |F(x) – 0| = |F(x)| ≤ |G(x)| < ε since
    limx→c G(x) = 0.
    Thus, limx→c F(x) = 0.

  25. Prove that limx→0 x4 sin2 (1/x) = 0. Hint: Use Problems 22 and 24.
    26. ❤
    PEMBAHASAN:
    For all x ≠ 0, 0 ≤ sin2 (1/x) ≤ 1 so
    x4 sin2 (1/x) ≤ x4 for all x ≠ 0. By Problem 18,
    limx→0 x4 = 0, so, by Problem 20,
    limx→0 x4 sin2 (1/x) = 0.

  26. Prove that limx→0+ √x = 0.
    27. ❤
    PEMBAHASAN:
    0 < x < δ ⇒ |√x – 0| = |√x| = √x < ε
    For x > 0, (√x)2 = x.
    √x < ε ⇔ (√x)2 = x < ε2
    δ = ε2; 0 < x < δ ⇒ √x < √δ = √ε2 = ε

  27. By considering left- and right-hand limits, prove that limx→0 |x| = 0.
    28. ❤
    PEMBAHASAN:
    limx→0+ |x| : 0 < x < δ ⇒ ||x| – 0| < ε
    For x ≥ 0, |x| = x.
    δ = ε; 0 < x < δ ⇒ ||x| – 0| = |x| = x < δ = ε
    Thus, limx→0+ |x| = 0.
    limx→0– |x| : 0 < 0 – x < δ ⇒ ||x| – 0| < ε
    For x < 0, |x| = –x; note also that ||x|| = |x| since |x| ≥ 0.
    δ = ε; 0 < –x < δ ⇒ ||x|| = |x| = –x < δ = ε
    Thus, limx→0– |x| = 0,
    since limx→0+ |x| = limx→0– |x| = 0, limx→0 |x| = 0.

  28. Prove that if |f(x)| < B for |x – a| < 1 and limx→a g(x) = 0, then limx→a f(x) g(x) = 0.
    29. ❤
    PEMBAHASAN:
    Choose ε > 0. Since limx→a g(x) = 0 there is some δ1 > 0 such that
    0 < |x – a| < δ1 ⇒ |g(x) – 0| < ε/B.
    Let δ = min{1, δ1}, then |f(x)| < B for |x – a| < δ or |x – a| < δ ⇒ |f(x)| < B.
    Thus, |x – a| < δ ⇒ |f(x) g(x) – 0| = |f(x) g(x)|
    = |f(x)| |g(x)| < B · ε/B = ε so limx→a f(x) g(x) = 0.

  29. Suppose that limx→a f(x) = L and that f(a) exists (though it may be different from L). Prove that f is bounded on some interval containing a; that is, show that there is an interval (c, d) with c < a < d and a constant M such that |f(x)| ≤ M for all x in (c, d).
    30. ❤
    PEMBAHASAN:
    Choose ε > 0. Since limx→a f(x) = L, there is a δ > 0 such that for 0 < |x – a| < δ, |f(x) – L| < ε.
    That is, for
    a – δ < x < a or a < x < a + δ.
    L – ε < f(x) < L + ε.
    Let f(a) = A,
    M = max {|L – ε|, |L + ε|, |A|}, c = a – δ, d = a + δ.
    Then for x in (c, d), |f(x)| ≤ M, since either x = a, in which case
    |f(x)| = |f(a)| = |A| ≤ M or 0 < |x – a| < δ so
    L – ε < f(x) < L – ε and |f(x)| < M.

  30. Prove that if f(x) ≤ g(x) for all x in some deleted interval about a and if limx→a f(x) = L and limx→a g(x) = M, then L ≤ M.
    31. ❤
    PEMBAHASAN:
    Suppose that L > M. Then L – M = α > 0.
    Now take ε < α/2 and δ = min{δ1, δ2} where
    0 < |x – a| < δ1 ⇒ |f(x) – L| < ε and
    0 < |x – a| < δ2 ⇒ |g(x) – L| < ε.
    Thus, for 0 < |x – a| < δ,
    L – ε < f(x) < L + ε and M – ε < g(x) < M + ε.
    Combine the inequalities and use the fact that f(x) ≤ g(x) to get
    L – ε < f(x) ≤ g(x) < M + ε which leads to
    L – ε < M + ε or L – M < 2ε.
    However,
    L – M = α > 2ε
    which is a contradiction.
    Thus L ≤ M.

  31. Which of the following are equivalent to the definition of limit?
    32. (a)
    For some ε > 0 and every δ > 0, 0 < |x – c| < δ ⇒ |f(x) – L| < ε.
    (b)
    For some δ > 0, there is a corresponding ε > 0 such that
     
    0 < |x – c| < ε ⇒ |f(x) – L| < δ
    (c)
    For every positive integer N, there is a corresponding positive integer M such that 0 < |x – c| < 1/M ⇒ |f(x) – L| < 1/N.
    (d)
    For every ε > 0, there is a corresponding δ > 0 such that 0 < |x – c| < δ and |f(x) – L| < ε for some x.
    ❤
    PEMBAHASAN:
    (b) and (c) are equivalent to the definition of limit.

  32. State in ε – δ language what it means to say limx→c f(x) ≠ L.
    33. ❤
    PEMBAHASAN:
    For every ε > 0 and δ > 0 there is some x with
    0 < |x – c| < δ such that |f(x) – L| > ε.

  33. Suppose we wish to give an ε – δ proof that
    34. lim
    x → 3
    x + 6
    = –1
    x4 – 4x3 – 2x2 + x + 6
    We begin by writing
    x + 6
    + 1 in the form (x – 3) g(x).
    x4 – 4x3 – 2x2 + x + 6
    (a)
    Determine g(x).
    (b)
    Could we choose δ = min(1, ε/n) for some n? Explain.
    (c)
    If we choose δ = min(¼, ε/m), what is the smallest integer m that we could use?
    ❤
    PEMBAHASAN:
    (a)
    g(x)
    =
    x3 – x2 – 2x – 4
    x4 – 4x3 + x2 + x + 6
    (b)
    No, because
    x + 6
    + 1
    x4 – 4x3 + x2 + x + 6
    has an asymptote at x ≈ 3.49.
    (c)
    If δ ≤ ¼, then 2.75 < x < 3 or 3 < x < 3.25 and by graphing
    y = |g(x)| =
    x3 – x2 – 2x – 4
     
     
    x4 – 4x3 + x2 + x + 6
     
     
    on the interval [2.75, 3.25], we see that
     
    0 <
    x3 – x2 – 2x – 4
    < 3
     
    x4 – 4x3 + x2 + x + 6
     
    so m mush be at least three.
     

Semoga Bermanfaat 😁

Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 - 1.2 Rigorous Study of Limits - Number 1 – 33. Silahkan untuk berkunjung kembali dikarenakan akan selalu ada update terbaru 😊😄🙏. Silahkan juga untuk memilih dan mendiskusikan di tempat postingan ini di kolom komentar ya supaya semakin bagus diskusi pada setiap postingan. Diperbolehkan request di kolom komentar pada postingan ini tentang rangkuman atau catatan atau soal dan yang lain atau bagian hal yang lainnya, yang sekiranya belum ada di website ini. Terima kasih banyak sebelumnya 👍. Semoga bermanfaat dan berkah untuk kita semua. Amiiinnn 👐👐👐
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