0

Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 1 - 1.1 Number 1 – 28

Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 Section 1.1

CHAPTER 1 LIMITS

SECTION 1.1 Introduction to Limits


Problem Set 1.1, Number 1 – 28.

Calculus 9th Purcell Chapter 1 - 1.1

    In Problems 1 – 6, find the indicated limit.
  1. lim
    (x – 5)
    x → 3
     
    2. ❤
    PEMBAHASAN:
    lim
    (x – 5)
    = –2
    x → 3
     
     

  2. lim
    (1 – 2t)
    t → –1
     
    3. ❤
    PEMBAHASAN:
    lim
    (1 – 2t)
    = 3
    t → –1
     
     

  3. lim
    (x2 + 2x – 1)
    x → –2
     
    4. ❤
    PEMBAHASAN:
    lim
    (x2 + 2x – 1)
    = (–2)2 + 2(–2) – 1 = –1
    x → –2
     
     

  4. lim
    (x2 + 2t – 1)
    x → –2
     
    5. ❤
    PEMBAHASAN:
    lim
    (x2 + 2t – 1)
    = (–2)2 + 2t – 1 = 3 + 2t
    x → –2
     
     

  5. lim
    (t2 – 1)
    t → –1
     
    6. ❤
    PEMBAHASAN:
    lim
    (t2 – 1)
    = ((–1)2 – 1) = 0
    t → –1
     
     

  6. lim
    (t2 – x2)
    t → –1
     
    7. ❤
    PEMBAHASAN:
    lim
    (t2 – x2)
    = ((–1)2 – x2) = 1 – x2
    t → –1
     
     

    Semoga Bermanfaat 😁

    In Problems 7-18, find the indicated limit. In most cases, it will be wise to do some algebra first (see Example 2).
  7.  
    lim
    x → 2
    x2 – 4
    x – 2
    8. ❤
    PEMBAHASAN:
     
     
    lim
    x → 2
    x2 – 4
    =
     
    lim
    x → 2
    (x – 2)(x + 2)
     
     
    x – 2
    x – 2
     
     
     
    =
    lim
    (x + 2)
     
    x → 2
     
     
     
    =
    2 + 2 = 4

  8.  
    lim
    t → –7
    t2 + 4t – 21
    t + 7
    9. ❤
    PEMBAHASAN:
     
    lim
    t → –7
    t2 + 4t – 21
    t + 7
    =
     
    lim
    t → –7
    (t + 7)(t – 3)
    t + 7
    =
    lim
    (t – 3)
    t → –7
     
    =
    –7 – 3 = –10

  9.  
    lim
    x → –1
    x3 – 4x2 + x + 6
    x + 1
    10. ❤
    PEMBAHASAN:
     
    lim
    x → –1
    x3 – 4x2 + x + 6
    x + 1
    =
     
    lim
    x → –1
    (x + 1)(x2 – 5x + 6)
    x + 1
    =
    lim
    (x2 – 5x + 6)
    x → –1
     
    =
    (–1)2 – 5(–1) + 6 = 12

  10.  
    lim
    x → 0
    x4 + 2x3 – x2
    x2
    11. ❤
    PEMBAHASAN:
     
    lim
    x → 0
    x4 + 2x3 – x2
    x2
    =
     
    lim
    x → 0
    (x2 + 2x – 1) = –1
     

  11.  
    lim
    x → –t
    x2 – t2
    x + t
    12. ❤
    PEMBAHASAN:
     
    lim
    x → –t
    x2 – t2
    x + t
    =
     
    lim
    x → –t
    (x + t)(x – t)
    x + t
    =
     
    lim
    x → –t
    (x – t)
     
    =
    –t – t = –2t

  12.  
    lim
    x → 3
    x2 – 9
    x – 3
    13. ❤
    PEMBAHASAN:
     
    lim
    x → 3
    x2 – 9
    x – 3
    =
     
    lim
    x → 3
    (x – 3)(x + 3)
    x – 3
    =
     
    lim
    x → 3
    (x + 3)
     
    =
    3 + 3 = 6

  13.  
    lim
    t → 2
    √[(t + 4)(t – 2)4]
    (3x – 6)2
    14. ❤
    PEMBAHASAN:
     
    lim
    t → 2
    √[(t + 4)(t – 2)4]
    (3x – 6)2
    =
     
    lim
    t → 2
    (t – 2)2 √(t + 4)
    9(t – 3)2
    =
     
    lim
    t → 2
    √(t + 4)
    9
    =
    √(2 + 4)
    9
    =
    √6
    9

  14.  
    lim
    t → 7+
    √(t – 7)3
    t – 7
    15. ❤
    PEMBAHASAN:
     
    lim
    t → 7+
    √(t – 7)3
    t – 7
    =
     
    lim
    t → 7+
    (t – 7) √(t – 7)
    (t – 7)
    =
     
    lim
    t → 7+
    √(t – 7)
     
    =
    √(7 – 7) = 0

  15.  
    lim
    x → 3
    x4 – 18x2 + 81
    (x – 3)2
    16. ❤
    PEMBAHASAN:
     
    lim
    x → 3
    x4 – 18x2 + 81
    (x – 3)2
    =
     
    lim
    x → 3
    (x2 – 9)2
    (x – 3)2
    =
     
    lim
    x → 3
    (x – 3)2 (x + 3)2
    (x – 3)2
    =
     
    lim
    x → 3
    (x + 3)2
     
    =
    (3 + 3)2 = 36

  16.  
    lim
    u → 1
    (3u + 4)(2u – 2)3
    (u – 1)2
    17. ❤
    PEMBAHASAN:
     
    lim
    u → 1
    (3u + 4)(2u – 2)3
    (u – 1)2
     
    lim
    u → 1
    8(3u + 4)(u – 1)3
    (u – 1)2
    =
     
    lim
    u → 1
    8(3u + 4)(u – 1)
     
    =
    8[3(1) + 4](1 – 1) = 0

  17.  
    lim
    h → 0
    (2 + h)2 – 4
    h
    18. ❤
    PEMBAHASAN:
     
    lim
    h → 0
    (2 + h)2 – 4
    h
    =
     
    lim
    h → 0
    4 + 4h + h2 – 4
    h
    =
     
    lim
    h → 0
    h2 + 4h
    h
    =
     
    lim
    h → 0
    (h + 4) = 4
     

  18.  
    lim
    h → 0
    (x + h)2 – x2
    h
    19. ❤
    PEMBAHASAN:
     
    lim
    h → 0
    (x + h)2 – x2
    h
    =
     
    lim
    h → 0
    x2 + 2xh + h2 – x2
    h
    =
     
    lim
    h → 0
    h2 + 2xh
    h
    =
     
    lim
    h → 0
    (h + 2x) = 2x
     

    Semoga Bermanfaat 😁

    In Problems 19-28, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point.
  19.  
    lim
    x → 0
    sin x
    2x
    20. ❤
    PEMBAHASAN:
    x
    (sin x)/(2x)
    1.
    0.420735
    0.1
    0.499167
    0.01
    0.499992
    0.001
    0.49999992
     
      
    –1.
    0.420735
    –0.1
    0.499167
    –0.01
    0.499992
    –0.001
    0.49999992
     
    lim
    x → 0
    sin x
    =
    0.5
    2x

  20.  
    lim
    t → 0
    1 – cos t
    2t
    21. ❤
    PEMBAHASAN:
    t
    (1 – cos t)/(2t)
    1.
    0.229849
    0.1
    0.0249792
    0.01
    0.00249998
    0.001
    0.00024999998
     
      
    –1.
    –0.229849
    –0.1
    –0.0249792
    –0.01
    –0.00249998
    –0.001
    –0.00024999998
     
    lim
    t → 0
    1 – cos t
    =
    0
    2t

  21.  
    lim
    x → 0
    (x – sin x)2
    x2
    22. ❤
    PEMBAHASAN:
    x
    (x – sin x)2/(x)2
    1.
    0.0251314
    0.1
    2.775 × 10−6
    0.01
    2.77775 × 10−10
    0.001
    2.77778 × 10−14
     
      
    –1.
    0.0251314
    –0.1
    2.775 × 10−6
    –0.01
    2.77775 × 10−10
    –0.001
    2.77778 × 10−14
     
    lim
    x → 0
    (x – sin x)2
    =
    0
    x2

  22.  
    lim
    x → 0
    (1 – cos x)2
    x2
    23. ❤
    PEMBAHASAN:
    x
    (1 – cos x)2/(x)2
    1.
    0.211322
    0.1
    0.00249584
    0.01
    0.0000249996
    0.001
    2.5 × 10−7
     
      
    –1.
    0.211322
    –0.1
    0.00249584
    –0.01
    0.0000249996
    –0.001
    2.5 × 10−7
     
    lim
    x → 0
    (1 – cos x)2
    =
    0
    x2

  23.  
    lim
    t → 1
    t2 – 1
    sin (t – 1)
    24. ❤
    PEMBAHASAN:
    t
    (t2 – 1)/(sin (t – 1))
    2.
    3.56519
    1.1
    2.1035
    1.01
    2.01003
    1.001
    2.001
     
      
    0
    1.1884
    0.9
    1.90317
    0.99
    1.99003
    0.999
    1.999
     
    lim
    t → 1
    t2 – 1
    =
    2
    sin (t – 1)

  24.  
    lim
    x → 3
    x – sin (x – 3) – 3
    x – 3
    25. ❤
    PEMBAHASAN:
    x
    (x – sin (x – 3) – 3)/(x – 3)
    4.
    0.158529
    3.1
    0.00166583
    3.01
    0.0000166666
    3.001
    1.66667 × 10−7
     
      
    2.
    0.158529
    2.9
    0.00166583
    2.99
    0.0000166666
    2.999
    1.66667 × 10−7
     
    lim
    x → 3
    x – sin (x – 3) – 3
    =
    0
    x – 3

  25.  
    lim
    x → π
    1 + sin (x – 3π/2)
    x – π
    26. ❤
    PEMBAHASAN:
    x
    (1 + sin (x – 3π/2))/(x – π)
    1. + π
    0.4597
    0.1 + π
    0.0500
    0.01 + π
    0.0050
    0.001 + π
    0.0005
     
      
    –1. + π
    –0.4597
    –0.1 + π
    –0.0500
    –0.01 + π
    –0.0050
    –0.001 + π
    –0.0005
     
    lim
    x → π
    1 + sin (x – 3π/2)
    =
    0
    x – π

  26.  
    lim
    t → 0
    1 – cot t
    1/t
    27. ❤
    PEMBAHASAN:
    t
    (1 – cot t)/(1/t)
    1.
    0.357907
    0.1
    –0.896664
    0.01
    –0.989967
    0.001
    –0.999
     
      
    –1.
    –1.64209
    –0.1
    –1.09666
    –0.01
    –1.00997
    –0.001
    –1.001
     
    lim
    t → 0
    1 – cot t
    =
    –1
    1/t

  27.  
    lim
    x → π/4
    (x – π/4)2
    (tan x – 1)2
    28. ❤
    PEMBAHASAN:
    x
    (x – π/4)^2/(tan x – 1)^2
    1. + π/4
    0.0320244
    0.1 + π/4
    0.201002
    0.01 + π/4
    0.245009
    0.001 + π/4
    0.2495
     
     
    –1. + π/4
    0.674117
    –0.1 + π/4
    0.300668
    –0.01 + π/4
    0.255008
    –0.001 + π/4
    0.2505
     
    lim
    x → π/4
    (x – π/4)2
    =
    0.25
    (tan x – 1)2

  28.  
    lim
    u → π/2
    2 – 2 sin u
    3u
    29. ❤
    PEMBAHASAN:
    u
    (2 – 2 sin u)/(3u)
    1. + π/2
    0.11921
    0.1 + π/2
    0.00199339
    0.01 + π/2
    0.0000210862
    0.001 + π/2
    2.12072 × 10−7
     
     
    –1. + π/2
    0.536908
    –0.1 + π/2
    0.00226446
    –0.01 + π/2
    0.0000213564
    –0.001 + π/2
    2.12342 × 10−7
     
    lim
    u → π/2
    2 – 2 sin u
    =
    0
    3u

Semoga Bermanfaat 😁

Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 1 - 1.1 Introduction to Limit - Number 1 – 28. 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 👐👐👐
Jangan lupa untuk 💏 SUBSCRIBE 👪 (Klik lonceng di bawah-kanan layar Anda) dan berikan komentar 💬 atau masukan serta share 👫 postingan ini ke teman-teman untuk berkembangnya https://www.bantalmateri.com/ ini 😀. Terima kasih dan semoga bermanfaat. 😋😆

Ahmad Qolfathiriyus Firdaus

We are bantalmateri.com that utilizes the internet and digital media in delivering material, questions and even the form of discussion. In the current generation, online learning methods (commonly called daring) are considered closer to students who are very integrated and difficult to separate from technology. The emergence of technology has also facilitated the implementation of schools even though students and educators alike have to adapt.

No comments:

Post a Comment