120 SOAL DAN PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI

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  120 Limit Fungsi Trigonometri
  SOAL DAN PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI
  2 tan −sin 1.
  Nilai lim = ⋯.
  cos → 3
  
3 A.
  D.
  √3 3√3
  
4
  5
  
1 B.
  E. √3 √3
  2
  
4
  3 C.
  √3
  2 Pembahasan
  2 tan − sin lim 2 tan 3 − sin
  3 = cos
  → 3
  cos 3 2. √3 − 12√3
  =
  1
  2 4√3 − √3
  2 =
  1
  2 4√3 − √3
  =
  1 = 3√3
  Jawaban A sin 2 2.
  Nilai lim = ⋯.
  sin + cos → 4 A.
  D. 0 √2
  1 B.
  E.
  −1 √2
  2 C.
  1 Pembahasan sin 2 lim sin 2. 4
  = sin + cos
  → 4
  sin 4 + cos
  4 sin 2 = sin 4 + cos
  4
  1 =
  1
  1 2 √2 + 2 √2
  1 =
  √2
  1 = 2 √2
  Jawaban B
2
  Pembahasan lim
  Jawaban D 4.
  Nilai lim
  → 4 cos 2 cos −sin = ⋯.
  A.
  −√2 D.
  
1
  
2
  √2 B. −
  1
  2
  √2 E.
  √2 C.
  → 4
  1 2 √2 + 1 2 √2
  cos 2 cos − sin = lim
  → 4
  2
  −
  2
  cos − sin = lim
  → 4
  (cos − sin )(cos + sin ) cos − sin = lim
  → 4
  (cos + sin ) = cos 4 + sin
  4 =
  1 2 √2 + 1 2 √2
  = √2
  = √2
  4 =
  120 Limit Fungsi Trigonometri 3.
  2
  Nilai lim
  → 4 1−2 2 cos −sin
  = ⋯.
  A.
  D.
  √2 B.
  1
  2
  √2 E.
  ∞ C.
  1 Pembahasan lim
  → 4
  1 − 2
  cos − sin = lim
  (cos + sin ) = cos 4 + sin
  → 4
  2
  − 2
  2
  cos − sin = lim
  → 4
  2
  −
  2
  cos − sin = lim
  → 4
  (cos − sin )(cos + sin ) cos − sin = lim
  → 4
  Jawaban E
  2 C.
  → 8
  A.
  D.
  1
  2
  √2 B.
  1
  2 E. 1 C.
  √2 Pembahasan lim
  → 8
  2
  2 −
  2
  2 sin 2 − cos 2 = lim
  (sin 2 − cos 2 )( 2 + cos 2 ) sin 2 − cos 2 = lim
  2 sin 2 −cos 2
  → 8
  ( 2 + cos 2 ) = ( 2.
  8 + cos 2. 8) = sin 4 + cos
  4 =
  1 2 √2 + 1 2 √2
  = √2
  Jawaban C 7.
  Nilai dari lim
  → 2 2 1−sin
  = ….
  A.
  −2 D.
  B.
  = ….
  → 8 2 2 − 2
  120 Limit Fungsi Trigonometri 5.
  → 4 sin 2 x+cos 2 x−2 sin .cos
sin −cos
  Nilai lim
  → 4 1−2 sin .cos sin −cos = ⋯.
  A.
  1 D. 0 B.
  1
  2
  √2 E.
  −1 C.
  1
  2 Pembahasan
  lim
  → 4
  1 − 2 sin . cos sin − cos = lim
  ;karena sin
  Nilai dari lim
  2
  x + cos
  2
  x = 1 = lim
  → 4
  (sin − cos )
  2
  sin − cos = lim
  → 4
  (sin − cos ) = sin 4 − cos
  4 =
  1 2 √2 − 1 2 √2
  =0
  Jawaban D 6.
– E.
  120 Limit Fungsi Trigonometri
  Pembahasan lim
  1 −
  2 =
  2 Jawaban D 8.
  Nilai lim
  → 4 sin −cos 1−tan = ⋯.
  A.
  −√2 D.
  
1
  
2
  √2 B. −
  1
  2
  √2 E.
  √2 C.
  → 4
  2 = 2 2
  sin − cos 1 − tan = lim
  → 4 sin −cos 1− sin cos
  ; karena tan =
  sin cos
  = lim
  → 4
  sin − cos cos x − sin cos
  = lim
  → 4
  cos (sin − cos ) cos x − sin = lim
  → 4
  − cos = − cos
  4 = −
  1 2 √2
  ;
  ;
  Pembahasan lim
  ) = lim
  → 2
  2
  1 − sin = lim
  → 2
  2
  2 (1 − sin ) .
  (1 + sin ) (1 + sin )
  = lim
  → 2
  2
  . (1 + sin )
  2
  (1 −
  2
  → 2
  1 =
  2
  . (1 + sin )
  2 .
  2
  = lim
  → 2
  (1 + sin )
  2
  = 2 (1 + sin 2)
  2
  2 = 2 (1 + 1)
  1
  2
  = 2 . 2
  Jawaban B
  120 Limit Fungsi Trigonometri sin 2 −2 sin 9.
  Nilai lim = ⋯. 3
  → 0
  3 A.
  D.
  −1
  2
  1 B.
  E.
  −2
  2
  1 C.
  −
  2 Pembahasan
  sin 2 − 2 sin 2 sin cos − 2 sin lim = lim
  3
  3 → 0 → 0
  2 sin (cos − 1) = lim
  3 → 0
  1
  2
  2 sin (−2 2 ) = lim
  3 → 0
  1 sin . sin 12 .sin
  2 = −4 lim
  3 → 0
  sin sin 12 sin 12 = −4 lim
  
→ 0 → 0 → 0
  lim . lim
  1
  1 = −4. 2 .
  2 =
  −1
  Jawaban D 2 2− 10.
  Nilai lim adalah ….
  1−cot → 4 A.
  D. 1
– 2 B.
  E. 2
– 1 C. Pembahasan
  2
  2 2 − (1 + ) 2 − lim = lim 1 − cot 1 − cot
  → 4 → 4
  2 1 − = lim
  1 − cot → 4
  (1 − cot )(1 + cot ) = lim 1 − cot
  → 4 (1 + cot ) = lim
  → 4 = (1 + cot 4) = 1 + 1 =2
  Jawaban E
  lim cos 4 x . sin x 120 Limit Fungsi Trigonometri  
  11.   
  Nilai x 5 x …  5   A. 3 B.
  1 3 C. 1 5 D. 5 E.
  Pembahasan
  cos 4 . sin sin lim = lim cos 4 . lim
  → 0 → 0 → 0
  5
  5
  1 = cos 4.0 .
  5
  1 = 1.
  5
  1 =
  5 Jawaban D
  sin 3 12.
  Nilai lim = ⋯.
  2 →0
  
2 A.
  D.
  3
  
3
  
1 B.
  E.
  2
  
2
  1 C.
  1
  2 Pembahasan
  sin 3
  3
  1 lim
  →0
  2 = 2 = 1
  2 Jawaban C
  sin 8 13.
  lim
  tan 2 →0
  2 A.
  D.
  4
  3
  1 B.
  E.
  3
  2 C.
  2 Pembahasan sin 8 sin 8 8 2 sin 8
  2
  8
  8 lim
  →0 →0 →0 →0
  tan 2 = lim tan 2 . 8 . 2 = lim 8 . lim tan 2 . 2 = 1.1. 2 = 4
  Jawaban A
  120 Limit Fungsi Trigonometri
  2 14.
  Nilai lim = ⋯.
  sin 4 →0
  1 A.
  D. 2
  4
  1 B.
  E. 4
  2 C.
  Pembahasan
  2
  2
  1 lim
  →0
  sin 4 = 4 =
  2 Jawaban B
  sin 2 15.
  Nilai lim =….
  sin 6 →0
  1 A.
  D. 3
  6
  1 B.
  E. 6
  3 C.
  2 Pembahasan sin 2
  2
  1 lim
  →0
  sin 6 = 6 =
  3 Jawaban B
  tan 5 16.
  Nilai lim = ⋯.
  sin 2 →0
  5
  2 A.
  D.
  2
  3
  3
  2 B.
  E.
  2
  5 C.
  2 Pembahasan tan 5 5 lim
  →0
  sin 2 =
  2 Jawaban A
  cos 2 17.
  Nilai lim = ⋯.
  sin →0
  A.
  D.1 −2 B.
  E.2 −1 C.
  Pembahasan cos 2 lim cos 2 = 1. cos 2.0 = 1.1 = 1
  →0 →0 →0
  sin = lim sin . lim
  Jawaban D
  120 Limit Fungsi Trigonometri 2 tan 18.
  Nilai lim = ⋯. 2
  →0 6
  1 A.
  D. 36
  3 B.
  E. 72
  3 C.
  12 Pembahasan 2 tan
  2
  2
  1 lim = lim . lim = . = 12.6 = 72
  1
  1
  →0 2 →0 →0
  6 tan 6 tan 6
  6
  6 Jawaban E
  tan 2 .tan 3 19.
  Nilai lim = ⋯. 2
  3 →0
  A.
  D. 2
  2 B.
  E. 6
  3
  3 C.
  2 Pembahasan
  tan 2 . tan 3 tan 2 tan 3
  2 lim = lim =
  2 →0 →0 →0
  3 3 . lim 3 . 3 = 2
  Jawaban D sin 4 −sin 2 20.
  Nilai lim = ⋯.
  6 →0
  1
  2 A.
  D.
  6
  3
  1 B.
  E.1
  3
  1 C.
  2 Pembahasan
  1 sin 4 − sin 2 2 cos 12(4 + 2 ).sin 2 (4 − 2 ) lim = lim
  →0 →0
  6
  6 2 cos 3 . sin = lim
  →0
  6 2 sin = cos 3 . lim
  →0 →0
  6 lim
  2
  = . cos 0.1
  6
  1 = 3 . 1.1
  1 =
  3 Jawaban B
  120 Limit Fungsi Trigonometri sin 5 +sin 21.
  Nilai lim = ⋯.
  6 →0
  1 A.
  D. −2
  2 B.
  E.1 −1
  1 C.
  3 Pembahasan
  sin 5 + sin sin 5 sin
  5
  1
  6 lim = lim
  →0 →0 →0
  6 6 + lim 6 = 6 + 6 = 6 = 1
  Jawaban E sin 7 +tan 3 −sin 5 22.
  Nilai lim = ⋯
  tan 9 −tan 3 −sin →0
  A.
  D. 3
  9 B.
  E. 1
  7 C.
  5 Pembahasan sin 7 + tan 3 − sin 5 7 + 3 − 5 5 lim
  →0
  tan 9 − tan 3 − sin = 9 − 3 − 1 = 5 = 1
  Jawaban E sin 4 +sin 2 23.
  Nilai lim = ⋯.
  3 cos →0
  A.
  D. 1,50 0,25 B.
  E. 2,00 0,50 C.
  1,00 Pembahasan sin 4 + sin 2 sin 4 + sin 2
  1 4 + 2
  1
  6
  1
  6 lim = lim . lim
  →0 →0 →0
  3 cos 3 cos = 3 . cos 0 = 3 . 1 = 3 = 2
  Jawaban E 4 cos 24.
  Nilai lim = ⋯.
  sin + sin 3 →0
  A.
  D. 1
  4
  3 B.
  E.
  3
  4
  4 C.
  3 Pembahasan
  4 cos
  4
  4
  4 lim cos =
  →0 →0 →0
  sin + sin 3 = lim sin + sin 3 . lim 1 + 3 . cos 0 = 4 . 1 = 1
  Jawaban D
  120 Limit Fungsi Trigonometri −sin 2 25.
  Nilai lim = ⋯.
sin 3 →0
  2
  2 A.
  D. −
  3
  3
  1
  3 B.
  E. −
  4
  4
  1 C.
  4 Pembahasan
  − sin 2 sin 2 lim −
  →0
sin 3
  = lim
  →0 sin 3
1 − sin 2
  = lim
  →0
  1 + sin 3 sin 2 lim 1 − lim
  →0 →0
  = sin 3 lim 1 + lim
  →0 →0
  1 − 2 = 1 + 3
  −1 =
  4 Jawaban B
  sin +tan 2 26.
  Nilai lim = ⋯.
  3 −sin 4 →0
  A.
  D. 3 −3 B.
  E.
  ∞ C.
  1 Pembahasan sin + tan 2 sin tan 2 lim
→0
  3 − sin 4 = lim
  →0
  3 sin 4 − sin tan 2 lim
lim
  →0 →0
  = sin 4 lim 3 − lim
  →0 →0
  1 + 2 = 3 − 4
  3 =
  −1 = −3
  Jawaban A 2 2 sin 4 . 3 +6 27.
  Nilai lim =…. 2
  2 +sin 3 . 2 →0
  A.
  D.
  5 B. E.
  3
  7 C.
  4
  120 Limit Fungsi Trigonometri
  Pembahasan
  2
  2
  2
  2 sin 4 . 3 + 6 sin 4 .
  3
6 lim
  2 →0
  2 + sin 3 . 2 = lim
  2 →0
  2 sin 3 . 2
2 sin 4 .
  3 lim + lim
  6
  
→0 →0
  = sin 3 . 2 lim 2 + lim
  →0 →0
  2
  lim 4 3 + lim
  6
  →0 . lim →0 →0
  = sin 3 lim 2 + lim cos 2
  . lim
  
→0 →0 →0
  2
  4. 0 + 0 = 0 + 3.1 =
  3 =0
  Jawaban A sin( + ) 3 28.
  Nilai lim = ⋯.
  ( + ) →− 3 3
  1 A.
  D. 1 −
  3
  1 B.
  E. 2 −
  2 C.
  Pembahasan Misalkan
  = ( + )3 Jika maka
  → − → 0
  3 sin( + ) 3 sin
  Jadi lim = lim = 1
  ( + ) →0 →− 3 3 Jawaban D sin sin( − )
  29. Jika lim = 1, maka lim = ⋯
  ( −1) →0 →1
  1 A.
  D.
  B.
  E.
  1
  2 C.
  Pembahasan sin( − ) sinπ( − 1) lim
  →1 →1
  ( − 1) = lim ( − 1) Misalkan
  ( − 1) = Jika
  → 1 maka → 0 sinπ( − 1) sinπ lim
  →1 →0
  = ( − 1) = = lim
  Jawaban C
  120 Limit Fungsi Trigonometri cos 3 .sin(12 −3 ) 30.
  Nilai lim = ⋯
  tan(4 − ) → 4
  3
  3 A.
  D. √3 − √2
  2
  2
  3
  3 B.
  E.
  − √2
  √3
  2
  2 C.
  Pembahasan cos 3 . sin(12 − 3 ) cos 3 . sin(12 − 3 ) lim = lim tan(4 − ) tan(4 − )
  → 4 → 4
  sin(12 − 3 ) = lim cos 3 lim tan(4 − )
  → 4 → 41 Untuk
  lim cos 3 = cos 3. = − √2
  4
  2 → 4 sin(12 −3 )
  Untuk lim
  tan(4 − ) → 4 Misalkan
  4 − = Jika maka
  → → 0 sehingga
  4
  sin(12 − 3 ) sin 3 lim
  →0
  tan(4 − ) = lim tan = 3
  → 4 sin(12 −3 )
  1
  3 Jadi
  lim cos 3 lim = − √2. 3 = − √2
  tan(4 − )
  2
  2 → → 4 4 Jawaban D 31.
  Nilai lim sin ( − ) tan ( + ) adalah ….
  4
  4 → 4 A.
  D.
  2 −1 B.
  E.
  1 −2 C.
  Pembahasan lim sin ( = lim sin ( 4 − ) tan ( 4 + )
  4 − ) cot ( 2 − ( 4 + ))
  → 4 → 4
  = lim sin ( 4 − ) cot ( 4 − )
  → 4
  cos ( 4 − ) = lim sin ( 4 − )
  → 4
  sin ( 4 − ) = lim cos ( 4 − )
  → 4
  = cos ( 4 − 4) = cos 0 = 1
  Jawaban B
  2 120 Limit Fungsi Trigonometri
  2 32.
  Nilai lim = ⋯. 2
  →0 A.
  D.
  1
  6 B. E.
  2
  8 C.
  4 Pembahasan
  2
  2 sin 2 sin 2 lim = lim
  2 →0 →0 →0
  . lim = 2.2 = 4 Atau dengan cara berikut
  2
  2
  2 sin 2
  2
  lim = (lim = 2 = 4
  2 →0 →0
  )
  Jawaban C 3 sin 2x 33.
  Nilai lim = ⋯. 31
  →0 tan x 2
  3
  6 A.
  D.
  2
  2
  4
  5 B.
  E.
  2
  25 C.
  2 Pembahasan
  3
  3
  3
  sin 2x sin 2
  2
  3
  2
  3
  6
  lim = (lim ) = ( ) = (2.2) = (2 ) = 2
  →0
  1 →0
  1
  3
  tan 2 x tan 12
  2 Jawaban D 2
  4 sin 2x 34.
  Nilai dari lim adalah ….
  tan 2 →0
  A.
  D. 4 −8 B.
  E. 8 −4 C.
  Pembahasan
  2
  4 sin 2x sin 2 sin 2 sin 2 sin 2
  2 lim
  →0 →0 →0 →0
  tan 2 = 4lim . tan 2 = 4 lim . lim tan 2 = 4.2. 2 = 8
  Jawaban E 21
  2 2 35.
  Nilai lim = ⋯.
  tan →0
  1 A.
  D. −2
  2 B.
  E. 1 −1
  1 C.
  −
  2 Pembahasan
  1
  2
  2
  1
  1
  1 2 sin 12 sin 12 lim
  →0 →0 →0
  tan = 2lim . lim tan = 2. 2 . 2 =
  2 Jawaban D
1 − 1 √31 − 1 √31 + 1 √2
4
41 + 1 = lim1) − 1 (√341 + 1) = lim41 + 1) = lim41 + 1) = lim41 + 1) = lim1 + 1) = lim41 + 1) =
  →0
  ) lim
  2
  5
  (√3
  2
  2
  2
  →0
  2
  1 (√2
  ) lim
  →0
  5
  (√3
  2
  2
  →0
  2
  )(√2
  5
  (√3
  2
  2
  1 (√2
  →0
  2
  →0
  1 (√2. 0
  sin + 1) .
  
→0
  . lim
  4
  4
  3
  →0
  2 (√lim
  2
  1 (√2
  . sin + 1) lim
  2 (√3
  4
  4
  2 (√3
  →0
  2
  1 (√2
  →0
  ) lim
  4
  4
  5
  →0
  2
  )(√2
  →0
  1 2 . 4 =
  1 2 .
  =
  tan 4√ √
  →0
  . lim
  sin 12 √
  →0
  1 2 lim
  sin 12 tan4√ 2 √ =
  lim
  5
  2 Pembahasan
  1
  4 E. 2 C.
  1
  8 D. 1 B.
  1
  A.
  = ⋯.
  →0 sin 1 2 tan 4√ 2 √
  Nilai lim
  120 Limit Fungsi Trigonometri 36.
  4
  4 = 1
  Jawaban D 37.
  = lim
  
2
  (2
  →0
  2
  2
  × √2
  5
  2
  √2
  →0
  5
  Nilai lim
  2
  √2
  →0
  Pembahasan lim
  √3 √4
  E. 1 C.
  2 B. √2 √3
  1
  D.
  A.
  →0 √2 2 +1−1 √3 5 + 4 = ….
  2
1 + 1)
  120 Limit Fungsi Trigonometri
  2
  1 = . (√3.0 + 1)
  (√1 + 1)
  2
  1 = 1 .
  2 = 1
  Jawaban E 4 38.
  Jika lim = 1, nilai yang memenuhi adalah …. 6
  →0 A.
  D. 4
  1 B.
  E. 5
  2 C.
  3 Pembahasan
  4
  4
  lim
  6
  2
  4
  2
  2 →0 →0 →0 →0 →0
  = lim . lim = lim . 1 = lim lim = 1 hanya terjadi jika nilai sama dengan pangkat dari sin , yaitu = 2 2
  →0 Jawaban B 3
  1− 39.
  Nilai lim =….
  .tan →0
  1 A.
  D. 2
  2 B.
  E. 3
  1
  3 C.
  2 Pembahasan
  3
  2
  1 − )
  (1 − cos )(1 + cos + lim = lim
  →0 →0
  . tan . tan (1 − cos )
  2
  = lim ) (1 + cos +
  →0 →0
  . tan lim
  1
  2 2 .
  2
  2
  = lim (1 + cos + )
  →0 →0
  . tan lim sin 12 sin 12
  2
  = 2lim lim ) (1 + cos +
  →0 →0 →0 →0
  lim tan . lim
  1
  1
  2
  = 2. 0) (1 + cos 0 + 2 . 2 .
  1
  2
  = 0) (1 + cos 0 +
  2
  1
  2
  = ) (1 + 1 + 1
  2
  3 =
  2 Jawaban C
  2 120 Limit Fungsi Trigonometri
  1− 40.
  Nilai lim = ⋯.
  tan →0
  A.
  D.
  3 −1 B.
  E.
  1 −3 C.
  Pembahasan
  2
  2
  1 − sin sin lim
  →0 →0 →0 →0
  tan = lim tan = lim . lim tan = 1.1 = 1
  Jawaban B .tan 2 41.
  Nilai lim = ⋯. 2
  1− →0
  1 A.
  D.
  2
  4
  1 B.
  E.
  4
  2 C.
  1 Pembahasan . tan 2 . tan 2 tan 2 lim
  2
  2 →0 →0 →0 →0
  1 − = lim = lim sin . lim sin = 1.2 = 2
  Jawaban C 2 1− 42.
  Nilai lim = ⋯. 2
  →0 tan( + ) 4
  1 A.
  D. −1
  √2
  2 B.
  E.
  √3 C.
  1 Pembahasan
  2
  2
  1 − lim = lim
  →0 →0
  2
  2
  tan ( + 4) tan ( + 4) sin sin
  1 = lim
  →0 →0 →0
  . lim . lim tan ( + 4) 1 = 1.1. tan (0 + 4)
  1 = 1.
  1 = 1
  Jawaban C 2 1− 43.
  Nilai lim = ⋯. 2
  →0 cot( − ) 3 A.
  D.
  1 −√2 B.
  E.
  −√31 C.
  − √3
  3
  120 Limit Fungsi Trigonometri
  A.
  lim
  2 Pembahasan
  1
  2 C.
  4 E.
  1
  D.1 B.
  4 2 −8 +4 = ⋯.
  1 −
  →1 1− 2 ( −1)
  Nilai lim
  Jawaban E 45.
  1 2 . 1
  1 2 .
  sin tan = −
  →0
  →1
  2
  →0
  1 4 lim
  1
  1 4 . 1.1 =
  =
  sin( − 1) ( − 1)
  →1
  sin( − 1) ( − 1) . lim
  →1
  − 2 + 1) =
  ( − 1)
  2
  ( − 1) 4(
  2
  →1
  − 8 + 4 = lim
  2
  4
  sin sin 2 . lim
  1 2 lim
  Pembahasan lim
  cot ( − 3) = lim
  1 cot (0 − 3) =
  1 cot ( − 3) = 1.1.
  →0
  sin . lim
  →0
  sin . lim
  →0
  2
  3 = −√3
  2
  →0
  cot ( − 3) = lim
  2
  2
  1 −
  →0
  1 cot (− 3) = − tan
  Jawaban E 44.
  2 sin 2 tan = −
  4 C.
  2
  −
  →0
  − 1 2 sin 2 tan = lim
  2
  →0
  −1 Pembahasan lim
  1
  Nilai lim
  −
  −2 E.
  2 B.
  1
  −
  −4 D.
  A.
  →0 2 −1 2 sin 2 tan = ⋯.
  4 Jawaban B
  2 120 Limit Fungsi Trigonometri x −4x+4 46.
  Nilai lim =…. 2
  1−cos (x−2) x→2
  1
  1 A.
  D. −
  4
  2 B.
  E. 1
  1 C.
  4 Pembahasan
  2
  2
  x − 4x + 4 ( − 2) lim = lim
  2
  2 x→2 x→2
  1 − cos sin (x − 2) (x − 2)
  ( − 2) ( − 2) = lim
  x→2 x→2
  sin(x − 2) . lim sin(x − 2) = 1.1 = 1
  Jawaban E 2 tan 3 47.
  Nilai lim = ⋯. 2
  1− →0
  A.
  D.6 B. E.12
  2 C.
  3 Pembahasan 2 . tan 3 2 . tan 3 2 tan 3 lim = lim
  2
  2 →0 →0 →0 →0
  1 − = lim sin . lim sin = 2.3 = 6
  Jawaban D 1− 48.
  Nilai lim = ⋯.
  sin →0
  A.
  D.1
  1 B.
  E. 2
  4
  1 C.
  2 Pembahasan
  1
  2
  2 1 −
  1
  1
  1 2 sin 12 lim sin
  →0 →0 →0 →0
  sin = lim sin = 2. lim sin . lim 2 = 2. 2 . sin 2 . 0 = 0
  Jawaban A 2 sin 3x 49.
  Nilai lim =….
  1−cos →0
  A.
  D. 12
  3 B.
  E. 18
  6 C.
  9 Pembahasan
  2
  2
  sin 3x sin 3x lim = lim
  →0 →0
  1 1 − cos
  2
  2
  2 1 sin 3 sin 3 = . lim
  →0 →0
  2 lim sin 12 sin 12
  120 Limit Fungsi Trigonometri
  →0 1−cos 8
  1 − cos 8
  →0
  2 Pembahasan lim
  1 E. 8 C.
  D. 4 B.
  A.
  4 2 =….
  Nilai lim
  2
  2 Jawaban E 51.
  1
  2 =
  1
  1 2 .
  12 = 2.
  →0
  12 . lim
  4
  = lim
  = 2 lim
  4
  1 2 . 4.4 = 8
  sin 4 =
  →0
  sin 4 . lim
  →0
  1 2 lim
  =
  2
  2
  →0
  →0
  2 4 lim
  =
  2
  4
  4
  2
  2
  →0
  2
  =
  2 = 18
  4 D.
  1
  −
  A.
  →0 1−cos 2 = ⋯.
  Nilai lim
  Jawaban E 50.
  1
  4 B.
  9
  2 =
  1
  3
  2 .
  1
  3
  1 2 .
  1
  −
  2
  2
  1
  2
  →0
  = 2 lim
  2
  2
  1
  2
  →0
  1
  = lim
  2
  1 − cos
  →0
  Pembahasan lim
  2 C.
  1
  2 E.
  Jawaban E
  120 Limit Fungsi Trigonometri 1− 2 52.
  Nilai lim =….
  1−cos 4 →0
  1 A.
  −
  2
  1 B.
  −
  4 C.
  1 D.
  16
  1 E.
  4 Pembahasan
  2
  1 − 2 2 sin x lim
  = lim
  2 →0
  1 − cos 4 x→0 2 sin 2x
  2
  sin x = lim
  2 x→0
  sin 2x sin sin = lim
  →0 →0
  sin 2 . lim sin 2
  1
  1 = 2 .
  2
  1 =
  4 Jawaban E
  1− 2 53.
  Nilai lim = ⋯.
  →0 A.
  D.2 −8 B.
  E.4 C.
  1 Pembahasan
  2
  1 − 2
  2 lim = lim
  →0 →0
  sin sin = 2 lim
  →0
  sin sin = 2 lim
  →0 →0
  . lim tan = 2.1.1 = 2
  Jawaban D (1−cos 4 ) sin 54.
  Nilai lim = ⋯. 2
  tan 3 →0 128
  8 A.
  D.
  3
  3
  32
  4 B.
  E.
  3
  3
  16 C.
  3 Pembahasan
  2
  (1 − cos 4 ) sin 2. 2 . sin lim
  = lim
  2
  2 →0 →0
  tan 3 tan 3
  120 Limit Fungsi Trigonometri
  2
  2 . sin = 2 lim
  2 →0
  tan 3
  
2
  sin 2 sin = 2 (lim . lim
  →0 →0
  ) tan 3
  1
  2 = 2. 2 .
  3
  8 =
  3 Jawaban D
  cos 4 −1 55.
  Nilai lim
  tan 2 →0
  A.
  D.
  4 −2 B.
  E.
  2 −4 C.
  −1 Pembahasan
  2
  cos 4 − 1 −2
  2 lim = lim
  →0
  tan 2 →0 tan 2
  2
  2 = −2 lim
  →0
  tan 2 sin 2 sin 2 = −2. lim
  →0 →0
  . lim tan 2
  2 = −2.2.
  2 = −4
  Jawaban E cos 6 −1 56. 1 Nilai lim
  →0 sin 2 A.
  D.
  36 −9 B.
  E.
  9 −36 C.
  Pembahasan
  2
  cos 6 − 1 −2
  3 lim = lim
  →0 →0
  sin 12 sin 12
  2
  3 = −2 lim
  →0
  sin 12 sin 3 sin 3 = −2. lim
  →0 →0
  . lim sin 12 3 = −2.3.
  1
 &

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