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![Muhammad Arif,S.Pd,M.Pd.
120 Limit Fungsi Trigonometri
22
SMAN 12 MAKASSAR
57. Nilai lim
𝑥→0
4x cos6𝑥 −4𝑥
(2𝑥)2.sin5𝑥
=
A. −
18
5
D. 2
B. −
5
18
E.
18
5
C.
5
18
Pembahasan
lim
𝑥→0
4x cos 6𝑥 − 4𝑥
(2𝑥)2. sin 5𝑥
= lim
𝑥→0
4𝑥(cos 6𝑥 − 1)
(2𝑥)2. sin 5𝑥
= lim
𝑥→0
4𝑥(−2 𝑠𝑖𝑛2
3𝑥)
(2𝑥)2. sin 5𝑥
= lim
𝑥→0
−8𝑥. sin 3𝑥 sin 3𝑥
(2𝑥)2. sin 5𝑥
= lim
𝑥→0
−8𝑥
sin 5𝑥
. lim
𝑥→0
sin 3𝑥
2𝑥
. lim
𝑥→0
sin 3𝑥
2𝑥
=
−8
5
.
3
2
.
3
2
= −
18
5
Jawaban A
58. Jika diketahui 𝑚 = lim
𝑥→0
cos 𝑥−1
cos2𝑥−1
dan 𝑛 = lim
𝑥→2
[
1
𝑥−2
−
4
𝑥2−4
], maka 𝑚 + 𝑛 =….
A. −1 D.
1
2
B. −
1
2
C. 1
C. 0
Pembahasan
𝑚 = lim
𝑥→0
cos 𝑥 − 1
cos 2𝑥 − 1
𝑚 = lim
𝑥→0
−2 𝑠𝑖𝑛2 1
2
𝑥
−2 𝑠𝑖𝑛2 𝑥
𝑚 = lim
𝑥→0
𝑠𝑖𝑛2 1
2
𝑥
𝑠𝑖𝑛2 𝑥
𝑚 = (lim
𝑥→0
𝑠𝑖𝑛
1
2
𝑥
𝑠𝑖𝑛𝑥
)
2
𝑚 = (
1
2
)
2
𝑚 =
1
4
𝑛 = lim
𝑥→2
[
1
𝑥 − 2
−
4
𝑥2 − 4
]
𝑛 = lim
𝑥→2
[
𝑥 + 2
𝑥2 − 4
−
4
𝑥2 − 4
]
𝑛 = lim
𝑥→2
[
𝑥 + 2 − 4
𝑥2 − 4
]
𝑛 = lim
𝑥→2
[
𝑥 − 2
𝑥2 − 4
]
𝑛 = lim
𝑥→2
𝑥 − 2
(𝑥 − 2)(𝑥 + 2)
𝑛 = lim
𝑥→2
1
(𝑥 + 2)
𝑛 =
1
2 + 2
𝑛 =
1
4
Nilai 𝑚 + 𝑛 =
1
4
+
1
4
=
1
2
Jawaban D](https://image.slidesharecdn.com/120soaldanpembahasanlimitfungsitrigonometri-190825125004/75/120-soal-dan-pembahasan-limit-fungsi-trigonometri-22-2048.jpg)
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120 soal dan pembahasan limit fungsi trigonometri
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120 soal dan pembahasan limit fungsi trigonometri
- 1. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 1 SMAN 12 MAKASSAR SOAL DAN PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI 1. Nilai lim 𝑥→ 𝜋 3 2 tan 𝑥−sin 𝑥 cos 𝑥 = ⋯. A. 3√3 D. 3 4 √3 B. 5 2 √3 E. 1 4 √3 C. 3 2 √3 Pembahasan lim 𝑥→ 𝜋 3 2 tan 𝑥 − sin 𝑥 cos 𝑥 = 2 tan 𝜋 3 − sin 𝜋 3 cos 𝜋 3 = 2. √3 − 1 2 √3 1 2 = 4√3 − √3 2 1 2 = 4√3 − √3 1 = 3√3 Jawaban A 2. Nilai lim 𝑥→ 𝜋 4 sin 2𝑥 sin 𝑥+ cos 𝑥 = ⋯. A. √2 D. 0 B. 1 2 √2 E. −1 C. 1 Pembahasan lim 𝑥→ 𝜋 4 sin 2𝑥 sin 𝑥 + cos 𝑥 = sin 2. 𝜋 4 sin 𝜋 4 + cos 𝜋 4 = sin 𝜋 2 sin 𝜋 4 + cos 𝜋 4 = 1 1 2 √2 + 1 2 √2 = 1 √2 = 1 2 √2 Jawaban B
- 2. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 2 SMAN 12 MAKASSAR 3. Nilai lim 𝑥→ 𝜋 4 1−2 𝑠𝑖𝑛2 𝑥 cos 𝑥−sin 𝑥 = ⋯. A. 0 D. √2 B. 1 2 √2 E. ∞ C. 1 Pembahasan lim 𝑥→ 𝜋 4 1 − 2 𝑠𝑖𝑛2 𝑥 cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 𝑐𝑜𝑠2 𝑥 + 𝑠𝑖𝑛2 𝑥 − 2 𝑠𝑖𝑛2 𝑥 cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 𝑐𝑜𝑠2 𝑥 − 𝑠𝑖𝑛2 𝑥 cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 (cos 𝑥 − sin 𝑥)(cos 𝑥 + sin 𝑥) cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 (cos 𝑥 + sin 𝑥) = cos 𝜋 4 + sin 𝜋 4 = 1 2 √2 + 1 2 √2 = √2 Jawaban D 4. Nilai lim 𝑥→ 𝜋 4 cos 2𝑥 cos 𝑥−sin 𝑥 = ⋯. A. −√2 D. 1 2 √2 B. − 1 2 √2 E. √2 C. 0 Pembahasan lim 𝑥→ 𝜋 4 cos 2𝑥 cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 𝑐𝑜𝑠2 𝑥 − 𝑠𝑖𝑛2 𝑥 cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 (cos 𝑥 − sin 𝑥)(cos 𝑥 + sin 𝑥) cos 𝑥 − sin 𝑥 = lim 𝑥→ 𝜋 4 (cos 𝑥 + sin 𝑥) = cos 𝜋 4 + sin 𝜋 4 = 1 2 √2 + 1 2 √2 = √2 Jawaban E
- 3. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 3 SMAN 12 MAKASSAR 5. Nilai lim 𝑥→ 𝜋 4 1−2 sin 𝑥.cos 𝑥 sin 𝑥−cos 𝑥 = ⋯. A. 1 D. 0 B. 1 2 √2 E. −1 C. 1 2 Pembahasan lim 𝑥→ 𝜋 4 1 − 2 sin 𝑥 . cos 𝑥 sin 𝑥 − cos 𝑥 = lim 𝑥→ 𝜋 4 sin2x+cos2x−2 sin 𝑥.cos 𝑥 sin 𝑥−cos 𝑥 ;karena sin2 x + cos2 x = 1 = lim 𝑥→ 𝜋 4 (sin 𝑥 − cos 𝑥)2 sin 𝑥 − cos 𝑥 = lim 𝑥→ 𝜋 4 (sin 𝑥 − cos 𝑥) = sin 𝜋 4 − cos 𝜋 4 = 1 2 √2 − 1 2 √2 =0 Jawaban D 6. Nilai dari lim 𝑥→ 𝜋 8 𝑠𝑖𝑛22𝑥−𝑐𝑜𝑠22𝑥 sin 2𝑥−cos 2𝑥 = …. A. 0 D. 1 2 √2 B. 1 2 E. 1 C. √2 Pembahasan lim 𝑥→ 𝜋 8 𝑠𝑖𝑛2 2𝑥 − 𝑐𝑜𝑠2 2𝑥 sin 2𝑥 − cos 2𝑥 = lim 𝑥→ 𝜋 8 (sin 2𝑥 − cos 2𝑥)(𝑠𝑖𝑛 2𝑥 + cos 2𝑥) sin 2𝑥 − cos 2𝑥 = lim 𝑥→ 𝜋 8 (𝑠𝑖𝑛 2𝑥 + cos 2𝑥) = (𝑠𝑖𝑛 2. 𝜋 8 + cos 2. 𝜋 8 ) = sin 𝜋 4 + cos 𝜋 4 = 1 2 √2 + 1 2 √2 = √2 Jawaban C 7. Nilai dari lim 𝑥→ 𝜋 2 𝑥 𝑐𝑜𝑡2 𝑥 1−sin 𝑥 = …. A. −2𝜋 D. 𝜋 B. – 𝜋 E. 2𝜋 C. 0
- 4. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 4 SMAN 12 MAKASSAR Pembahasan lim 𝑥→ 𝜋 2 𝑥 𝑐𝑜𝑡2 𝑥 1 − sin 𝑥 = lim 𝑥→ 𝜋 2 𝑥 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠2 𝑥(1 − sin 𝑥) . (1 + sin 𝑥) (1 + sin 𝑥) = lim 𝑥→ 𝜋 2 𝑥 𝑐𝑜𝑠2 𝑥. (1 + sin 𝑥) 𝑠𝑖𝑛2 𝑥(1 − 𝑠𝑖𝑛2 𝑥) = lim 𝑥→ 𝜋 2 𝑥 𝑐𝑜𝑠2 𝑥. (1 + sin 𝑥) 𝑠𝑖𝑛2 𝑥. 𝑐𝑜𝑠2 𝑥 = lim 𝑥→ 𝜋 2 𝑥 (1 + sin 𝑥) 𝑠𝑖𝑛2 𝑥 = 𝜋 2 (1 + sin 𝜋 2 ) 𝑠𝑖𝑛2 𝜋 2 = 𝜋 2 (1 + 1) 12 = 𝜋 2 . 2 1 = 𝜋 ;𝑐𝑜𝑡2 𝑥 = 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠2 𝑥 ;1 − 𝑠𝑖𝑛2 𝑥 = 𝑐𝑜𝑠2 𝑥 Jawaban D 8. Nilai lim 𝑥→ 𝜋 4 sin 𝑥 −cos 𝑥 1−tan 𝑥 = ⋯. A. −√2 D. 1 2 √2 B. − 1 2 √2 E. √2 C. 0 Pembahasan lim 𝑥→ 𝜋 4 sin 𝑥 − cos 𝑥 1 − tan 𝑥 = lim 𝑥→ 𝜋 4 sin 𝑥 −cos 𝑥 1− sin 𝑥 cos 𝑥 ; karena tan 𝑥 = sin 𝑥 cos 𝑥 = lim 𝑥→ 𝜋 4 sin 𝑥 − cos 𝑥 cos x − sin 𝑥 cos 𝑥 = lim 𝑥→ 𝜋 4 cos 𝑥 (sin 𝑥 − cos 𝑥) cos x − sin 𝑥 = lim 𝑥→ 𝜋 4 − cos 𝑥 = − cos 𝜋 4 = − 1 2 √2 Jawaban B
- 5. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 5 SMAN 12 MAKASSAR 9. Nilai lim 𝑥→ 0 sin 2𝑥 −2 sin 𝑥 𝑥3 = ⋯. A. 3 2 D. −1 B. 1 2 E. −2 C. − 1 2 Pembahasan lim 𝑥→ 0 sin 2𝑥 − 2 sin 𝑥 𝑥3 = lim 𝑥→ 0 2 sin 𝑥 cos 𝑥 − 2 sin 𝑥 𝑥3 = lim 𝑥→ 0 2 sin 𝑥 (cos 𝑥 − 1) 𝑥3 = lim 𝑥→ 0 2 sin 𝑥 (−2𝑠𝑖𝑛2 1 2 𝑥) 𝑥3 = −4 lim 𝑥→ 0 sin 𝑥 . sin 1 2 𝑥 . sin 1 2 𝑥 𝑥3 = −4 lim 𝑥→ 0 sin 𝑥 𝑥 lim 𝑥→ 0 sin 1 2 𝑥 𝑥 . lim 𝑥→ 0 sin 1 2 𝑥 𝑥 = −4. 1 2 . 1 2 =−1 Jawaban D 10. Nilai lim 𝑥→ 𝜋 4 2−𝑐𝑠𝑐2 𝑥 1−cot 𝑥 adalah …. A. – 2 D. 1 B. – 1 E. 2 C. 0 Pembahasan lim 𝑥→ 𝜋 4 2 − 𝑐𝑠𝑐2 𝑥 1 − cot 𝑥 = lim 𝑥→ 𝜋 4 2 − (1 + 𝑐𝑜𝑡2 𝑥) 1 − cot 𝑥 = lim 𝑥→ 𝜋 4 1 − 𝑐𝑜𝑡2 𝑥 1 − cot 𝑥 = lim 𝑥→ 𝜋 4 (1 − cot 𝑥)(1 + cot 𝑥) 1 − cot 𝑥 = lim 𝑥→ 𝜋 4 (1 + cot 𝑥) = (1 + cot 𝜋 4 ) = 1 + 1 =2 Jawaban E
- 6. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 6 SMAN 12 MAKASSAR 11. Nilai x xx x 5 sin.4cos 0 lim … A. 3 5 B. 1 C. 5 3 D. 5 1 E. 0 Pembahasan lim 𝑥→ 0 cos 4𝑥. sin 𝑥 5𝑥 = lim 𝑥→ 0 cos 4𝑥 . lim 𝑥→ 0 sin 𝑥 5𝑥 = cos 4.0 . 1 5 = 1. 1 5 = 1 5 Jawaban D 12. Nilai lim 𝑥→0 sin 3𝑥 2𝑥 = ⋯. A. 3 D. 2 3 B. 2 E. 1 2 C. 1 1 2 Pembahasan lim 𝑥→0 sin 3𝑥 2𝑥 = 3 2 = 1 1 2 Jawaban C 13. lim 𝑥→0 sin 8𝑥 tan 2𝑥 A. 4 D. 2 3 B. 3 E. 1 2 C. 2 Pembahasan lim 𝑥→0 sin 8𝑥 tan 2𝑥 = lim 𝑥→0 sin 8𝑥 tan 2𝑥 . 8𝑥 8𝑥 . 2𝑥 2𝑥 = lim 𝑥→0 sin 8𝑥 8𝑥 . lim 𝑥→0 2𝑥 tan 2𝑥 . 8𝑥 2𝑥 = 1.1. 8 2 = 4 Jawaban A
- 7. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 7 SMAN 12 MAKASSAR 14. Nilai lim 𝜃→0 2𝜃 sin4𝜃 = ⋯. A. 1 4 D. 2 B. 1 2 E. 4 C. 0 Pembahasan lim 𝜃→0 2𝜃 sin 4𝜃 = 2 4 = 1 2 Jawaban B 15. Nilai lim 𝑥→0 sin 2𝑥 sin 6𝑥 =…. A. 1 6 D. 3 B. 1 3 E. 6 C. 2 Pembahasan lim 𝑥→0 sin 2𝑥 sin 6𝑥 = 2 6 = 1 3 Jawaban B 16. Nilai lim 𝜃→0 tan 5𝜃 sin 2𝜃 = ⋯. A. 5 2 D. 2 3 B. 3 2 E. 2 5 C. 2 Pembahasan lim 𝜃→0 tan 5𝜃 sin 2𝜃 = 5 2 Jawaban A 17. Nilai lim 𝑥→0 𝑥 cos2𝑥 sin 𝑥 = ⋯. A. −2 D.1 B. −1 E.2 C. 0 Pembahasan lim 𝑥→0 𝑥 cos 2𝑥 sin 𝑥 = lim 𝑥→0 𝑥 sin 𝑥 . lim 𝑥→0 cos 2𝑥 = 1. cos 2.0 = 1.1 = 1 Jawaban D
- 8. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 8 SMAN 12 MAKASSAR 18. Nilai lim 𝑥→0 2𝑥 tan 𝑥 𝑡𝑎𝑛2 𝑥 6 = ⋯. A. 1 3 D. 36 B. 3 E. 72 C. 12 Pembahasan lim 𝑥→0 2𝑥 tan 𝑥 𝑡𝑎𝑛2 𝑥 6 = lim 𝑥→0 2𝑥 tan 𝑥 6 . lim 𝑥→0 𝑡𝑎𝑛 𝑥 tan 𝑥 6 = 2 1 6 . 1 1 6 = 12.6 = 72 Jawaban E 19. Nilai lim 𝑥→0 tan 2𝑥.tan 3𝑥 3𝑥2 = ⋯. A. 0 D. 2 B. 2 3 E. 6 C. 3 2 Pembahasan lim 𝑥→0 tan 2𝑥 . tan 3𝑥 3𝑥2 = lim 𝑥→0 tan 2𝑥 3𝑥 . lim 𝑥→0 tan 3𝑥 𝑥 = 2 3 . 3 = 2 Jawaban D 20. Nilai lim 𝑥→0 sin 4𝑥−sin2𝑥 6𝑥 = ⋯. A. 1 6 D. 2 3 B. 1 3 E.1 C. 1 2 Pembahasan lim 𝑥→0 sin 4𝑥 − sin 2𝑥 6𝑥 = lim 𝑥→0 2 cos 1 2 (4𝑥 + 2𝑥) . sin 1 2 (4𝑥 − 2𝑥) 6𝑥 = lim 𝑥→0 2 cos 3𝑥 . sin 𝑥 6𝑥 = 2 6 lim 𝑥→0 cos 3𝑥 . lim 𝑥→0 sin 𝑥 𝑥 = 2 6 . cos 0.1 = 1 3 . 1.1 = 1 3 Jawaban B
- 9. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 9 SMAN 12 MAKASSAR 21. Nilai lim 𝑥→0 sin 5𝑥+sin 𝑥 6𝑥 = ⋯. A. −2 D. 1 2 B. −1 E.1 C. 1 3 Pembahasan lim 𝑥→0 sin 5𝑥 + sin 𝑥 6𝑥 = lim 𝑥→0 sin 5𝑥 6𝑥 + lim 𝑥→0 sin 𝑥 6𝑥 = 5 6 + 1 6 = 6 6 = 1 Jawaban E 22. Nilai lim 𝑥→0 sin 7𝑥 +tan3𝑥−sin5𝑥 tan 9𝑥−tan 3𝑥 −sin 𝑥 = ⋯ A. 9 D. 3 B. 7 E. 1 C. 5 Pembahasan lim 𝑥→0 sin 7𝑥 + tan 3𝑥 − sin 5𝑥 tan 9𝑥 − tan 3𝑥 − sin 𝑥 = 7 + 3 − 5 9 − 3 − 1 = 5 5 = 1 Jawaban E 23. Nilai lim 𝑥→0 sin 4𝑥+sin2𝑥 3𝑥 cos 𝑥 = ⋯. A. 0,25 D. 1,50 B. 0,50 E. 2,00 C. 1,00 Pembahasan lim 𝑥→0 sin 4𝑥 + sin 2𝑥 3𝑥 cos 𝑥 = lim 𝑥→0 sin 4𝑥 + sin 2𝑥 3𝑥 . lim 𝑥→0 1 cos 𝑥 = 4 + 2 3 . 1 cos 0 = 6 3 . 1 1 = 6 3 = 2 Jawaban E 24. Nilai lim 𝑥→0 4𝑥 cos 𝑥 sin 𝑥+ sin3𝑥 = ⋯. A. 4 D. 1 B. 3 E. 3 4 C. 4 3 Pembahasan lim 𝑥→0 4𝑥 cos 𝑥 sin 𝑥 + sin 3𝑥 = lim 𝑥→0 4𝑥 sin 𝑥 + sin 3𝑥 . lim 𝑥→0 cos 𝑥 = 4 1 + 3 . cos 0 = 4 4 . 1 = 1 Jawaban D
- 10. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 10 SMAN 12 MAKASSAR 25. Nilai lim 𝑥→0 𝑥−sin 2𝑥 𝑥+sin 3𝑥 = ⋯. A. − 2 3 D. 2 3 B. − 1 4 E. 3 4 C. 1 4 Pembahasan lim 𝑥→0 𝑥 − sin 2𝑥 𝑥 + sin 3𝑥 = lim 𝑥→0 𝑥 𝑥 − sin 2𝑥 𝑥 𝑥 𝑥 + sin 3𝑥 𝑥 = lim 𝑥→0 1 − sin 2𝑥 𝑥 1 + sin 3𝑥 𝑥 = lim 𝑥→0 1 − lim 𝑥→0 sin 2𝑥 𝑥 lim 𝑥→0 1 + lim 𝑥→0 sin 3𝑥 𝑥 = 1 − 2 1 + 3 = −1 4 Jawaban B 26. Nilai lim 𝑥→0 sin 𝑥+tan 2𝑥 3𝑥−sin 4𝑥 = ⋯. A. −3 D. 3 B. 0 E. ∞ C. 1 Pembahasan lim 𝑥→0 sin 𝑥 + tan 2𝑥 3𝑥 − sin 4𝑥 = lim 𝑥→0 sin 𝑥 𝑥 + tan 2𝑥 𝑥 3𝑥 𝑥 − sin 4𝑥 𝑥 = lim 𝑥→0 sin 𝑥 𝑥 + lim 𝑥→0 tan 2𝑥 𝑥 lim 𝑥→0 3 − lim 𝑥→0 sin 4𝑥 𝑥 = 1 + 2 3 − 4 = 3 −1 = −3 Jawaban A 27. Nilai lim 𝑥→0 sin 4𝑥.𝑡𝑎𝑛23𝑥+6𝑥2 2𝑥2+sin3𝑥.𝑐𝑜𝑠2𝑥 =…. A. 0 D. 5 B. 3 E. 7 C. 4
- 11. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 11 SMAN 12 MAKASSAR Pembahasan lim 𝑥→0 sin 4𝑥 . 𝑡𝑎𝑛2 3𝑥 + 6𝑥2 2𝑥2 + sin 3𝑥 . 𝑐𝑜𝑠2𝑥 = lim 𝑥→0 sin 4𝑥 . 𝑡𝑎𝑛2 3𝑥 𝑥 + 6𝑥2 𝑥 2𝑥2 𝑥 + sin 3𝑥 . 𝑐𝑜𝑠2𝑥 𝑥 = lim 𝑥→0 sin 4𝑥 . 𝑡𝑎𝑛2 3𝑥 𝑥 + lim 𝑥→0 6𝑥 lim 𝑥→0 2𝑥 + lim 𝑥→0 sin 3𝑥 . 𝑐𝑜𝑠2𝑥 𝑥 = lim 𝑥→0 𝑠𝑖𝑛4𝑥 𝑥 . lim 𝑥→0 𝑡𝑎𝑛2 3𝑥 + lim 𝑥→0 6𝑥 lim 𝑥→0 2𝑥 + lim 𝑥→0 sin 3𝑥 𝑥 . lim 𝑥→0 cos 2𝑥 = 4. 02 + 0 0 + 3.1 = 0 3 =0 Jawaban A 28. Nilai lim 𝑥→− 𝜋 3 sin(𝑥+ 𝜋 3 ) (𝑥+ 𝜋 3 ) = ⋯. A. − 1 3 D. 1 B. − 1 2 E. 2 C. 0 Pembahasan Misalkan 𝑦 = (𝑥 + 𝜋 3 ) Jika 𝑥 → − 𝜋 3 maka 𝑦 → 0 Jadi lim 𝑥→− 𝜋 3 sin(𝑥+ 𝜋 3 ) (𝑥+ 𝜋 3 ) = lim 𝑦→0 sin 𝑦 𝑦 = 1 Jawaban D 29. Jika lim 𝑥→0 sin 𝑥 𝑥 = 1, maka lim 𝑥→1 sin(𝜋𝑥−𝜋) (𝑥−1) = ⋯ A. 0 D. 1 𝜋 B. 1 E. 𝜋 2 C. 𝜋 Pembahasan lim 𝑥→1 sin(𝜋𝑥 − 𝜋) (𝑥 − 1) = lim 𝑥→1 sinπ(𝑥 − 1) (𝑥 − 1) Misalkan (𝑥 − 1) = 𝑦 Jika 𝑥 → 1 maka 𝑦 → 0 lim 𝑥→1 sinπ(𝑥 − 1) (𝑥 − 1) = = lim 𝑦→0 sinπ 𝑦 𝑦 = 𝜋 Jawaban C
- 12. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 12 SMAN 12 MAKASSAR 30. Nilai lim 𝑥→ 𝜋 4 cos3𝑥.sin(12𝑥−3𝜋) tan(4𝑥−𝜋) = ⋯ A. 3 2 √3 D. − 3 2 √2 B. 3 2 √2 E. − 3 2 √3 C. 0 Pembahasan lim 𝑥→ 𝜋 4 cos 3𝑥 . sin(12𝑥 − 3𝜋) tan(4𝑥 − 𝜋) = lim 𝑥→ 𝜋 4 cos 3𝑥 . sin(12𝑥 − 3𝜋) tan(4𝑥 − 𝜋) = lim 𝑥→ 𝜋 4 cos 3𝑥 lim 𝑥→ 𝜋 4 sin(12𝑥 − 3𝜋) tan(4𝑥 − 𝜋) Untuk lim 𝑥→ 𝜋 4 cos 3𝑥 = cos 3. 𝜋 4 = − 1 2 √2 Untuk lim 𝑥→ 𝜋 4 sin(12𝑥−3𝜋) tan(4𝑥−𝜋) Misalkan 4𝑥 − 𝜋 = 𝑦 Jika 𝑥 → 𝜋 4 maka 𝑦 → 0 sehingga lim 𝑥→ 𝜋 4 sin(12𝑥 − 3𝜋) tan(4𝑥 − 𝜋) = lim 𝑦→0 sin 3𝑦 tan 𝑦 = 3 Jadi lim 𝑥→ 𝜋 4 cos 3𝑥 lim 𝑥→ 𝜋 4 sin(12𝑥−3𝜋) tan(4𝑥−𝜋) = − 1 2 √2. 3 = − 3 2 √2 Jawaban D 31. Nilai lim 𝑥→ 𝜋 4 sin ( 𝜋 4 − 𝑥) tan ( 𝜋 4 + 𝑥) adalah …. A. 2 D. −1 B. 1 E. −2 C. 0 Pembahasan lim 𝑥→ 𝜋 4 sin ( 𝜋 4 − 𝑥) tan ( 𝜋 4 + 𝑥) = lim 𝑥→ 𝜋 4 sin ( 𝜋 4 − 𝑥) cot ( 𝜋 2 − ( 𝜋 4 + 𝑥)) = lim 𝑥→ 𝜋 4 sin ( 𝜋 4 − 𝑥) cot ( 𝜋 4 − 𝑥) = lim 𝑥→ 𝜋 4 sin ( 𝜋 4 − 𝑥) cos ( 𝜋 4 − 𝑥) sin ( 𝜋 4 − 𝑥) = lim 𝑥→ 𝜋 4 cos ( 𝜋 4 − 𝑥) = cos ( 𝜋 4 − 𝜋 4 ) = cos 0 = 1 Jawaban B
- 13. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 13 SMAN 12 MAKASSAR 32. Nilai lim 𝑥→0 𝑠𝑖𝑛22𝑥 𝑥2 = ⋯. A. 1 D. 6 B. 2 E. 8 C. 4 Pembahasan lim 𝑥→0 𝑠𝑖𝑛2 2𝑥 𝑥2 = lim 𝑥→0 sin 2𝑥 𝑥 . lim 𝑥→0 sin 2𝑥 𝑥 = 2.2 = 4 Atau dengan cara berikut lim 𝑥→0 𝑠𝑖𝑛2 2𝑥 𝑥2 = (lim 𝑥→0 sin 2𝑥 𝑥 ) 2 = 22 = 4 Jawaban C 33. Nilai lim 𝑥→0 sin32x tan31 2 x = ⋯. A. 23 D. 26 B. 24 E. 25 C. 25 Pembahasan lim 𝑥→0 sin3 2x tan3 1 2 x = (lim 𝑥→0 sin 2𝑥 tan 1 2 𝑥 ) 3 = ( 2 1 2 ) 3 = (2.2)3 = (22)3 = 26 Jawaban D 34. Nilai dari lim 𝑥→0 4 sin22x 𝑥 tan 2𝑥 adalah …. A. −8 D. 4 B. −4 E. 8 C. 0 Pembahasan lim 𝑥→0 4 sin2 2x 𝑥 tan 2𝑥 = 4lim 𝑥→0 sin 2𝑥 sin 2𝑥 𝑥. tan 2𝑥 = 4 lim 𝑥→0 sin 2𝑥 𝑥 . lim 𝑥→0 sin 2𝑥 tan 2𝑥 = 4.2. 2 2 = 8 Jawaban E 35. Nilai lim 𝑥→0 2 𝑠𝑖𝑛21 2 𝑥 𝑥 tan 𝑥 = ⋯. A. −2 D. 1 2 B. −1 E. 1 C. − 1 2 Pembahasan lim 𝑥→0 2 𝑠𝑖𝑛2 1 2 𝑥 𝑥 tan 𝑥 = 2lim 𝑥→0 sin 1 2 𝑥 𝑥 . lim 𝑥→0 sin 1 2 𝑥 tan 𝑥 = 2. 1 2 . 1 2 = 1 2 Jawaban D
- 14. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 14 SMAN 12 MAKASSAR 36. Nilai lim 𝑥→0 sin 1 2 𝑥 tan 4√ 𝑥 2𝑥√ 𝑥 = ⋯. A. 1 8 D. 1 B. 1 4 E. 2 C. 1 2 Pembahasan lim 𝑥→0 sin 1 2 𝑥 tan 4√ 𝑥 2𝑥√ 𝑥 = 1 2 lim 𝑥→0 sin 1 2 𝑥 𝑥√ 𝑥 . lim 𝑥→0 tan 4√ 𝑥 √ 𝑥 = 1 2 . 1 2 . 4 = 4 4 = 1 Jawaban D 37. Nilai lim 𝑥→0 √2𝑥2+1−1 √3 𝑠𝑖𝑛5 𝑥+𝑥4 = …. A. 0 D. 1 2 B. √2 √3 E. 1 C. √3 √4 Pembahasan lim 𝑥→0 √2𝑥2 + 1 − 1 √3 𝑠𝑖𝑛5 𝑥 + 𝑥4 = lim 𝑥→0 √2𝑥2 + 1 − 1 √3 𝑠𝑖𝑛5 𝑥 + 𝑥4 × √2𝑥2 + 1 + 1 √2𝑥2 + 1 + 1 = lim 𝑥→0 (2𝑥2 + 1) − 1 (√3 𝑠𝑖𝑛5 𝑥 + 𝑥4)(√2𝑥2 + 1 + 1) = lim 𝑥→0 2𝑥2 (√3 𝑠𝑖𝑛5 𝑥 + 𝑥4)(√2𝑥2 + 1 + 1) = lim 𝑥→0 2𝑥2 (√3 𝑠𝑖𝑛5 𝑥 + 𝑥4) lim 𝑥→0 1 (√2𝑥2 + 1 + 1) = lim 𝑥→0 2𝑥2 𝑥2 ( √3 𝑠𝑖𝑛5 𝑥 + 𝑥4 𝑥2 ) lim 𝑥→0 1 (√2𝑥2 + 1 + 1) = lim 𝑥→0 2 (√3 𝑠𝑖𝑛5 𝑥 𝑥4 + 𝑥4 𝑥4) lim 𝑥→0 1 (√2𝑥2 + 1 + 1) = lim 𝑥→0 2 (√3 𝑠𝑖𝑛4 𝑥 𝑥4 . sin 𝑥 + 1) lim 𝑥→0 1 (√2𝑥2 + 1 + 1) = 2 (√lim 𝑥→0 3 𝑠𝑖𝑛4 𝑥 𝑥4 . lim 𝑥→0 sin 𝑥 + 1) . 1 (√2. 02 + 1 + 1)
- 15. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 15 SMAN 12 MAKASSAR = 2 (√3.0 + 1) . 1 (√1 + 1) = 2 1 . 1 2 = 1 Jawaban E 38. Jika lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛4 𝑥 𝑠𝑖𝑛6 𝑥 = 1, nilai 𝑎 yang memenuhi adalah …. A. 1 D. 4 B. 2 E. 5 C. 3 Pembahasan lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛4 𝑥 𝑠𝑖𝑛6 𝑥 = lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛2 𝑥 . lim 𝑥→0 𝑠𝑖𝑛4 𝑥 𝑠𝑖𝑛4 𝑥 = lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛2 𝑥 . 1 = lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛2 𝑥 lim 𝑥→0 𝑥 𝑎 𝑠𝑖𝑛2 𝑥 = 1 hanya terjadi jika nilai 𝑎 sama dengan pangkat dari sin 𝑥, yaitu 𝑎 = 2 Jawaban B 39. Nilai lim 𝑥→0 1−𝑐𝑜𝑠3 𝑥 𝑥.tan 𝑥 =…. A. 1 2 D. 2 B. 1 E. 3 C. 3 2 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠3 𝑥 𝑥. tan 𝑥 = lim 𝑥→0 (1 − cos 𝑥)(1 + cos 𝑥 + 𝑐𝑜𝑠2 𝑥) 𝑥. tan 𝑥 = lim 𝑥→0 (1 − cos 𝑥) 𝑥. tan 𝑥 lim 𝑥→0 (1 + cos 𝑥 + 𝑐𝑜𝑠2 𝑥) = lim 𝑥→0 2 . 𝑠𝑖𝑛2 1 2 𝑥 𝑥. tan 𝑥 lim 𝑥→0 (1 + cos 𝑥 + 𝑐𝑜𝑠2 𝑥) = 2lim 𝑥→0 sin 1 2 𝑥 𝑥 lim 𝑥→0 sin 1 2 𝑥 tan 𝑥 . lim 𝑥→0 lim 𝑥→0 (1 + cos 𝑥 + 𝑐𝑜𝑠2 𝑥) = 2. 1 2 . 1 2 . (1 + cos 0 + 𝑐𝑜𝑠2 0) = 1 2 (1 + cos 0 + 𝑐𝑜𝑠2 0) = 1 2 (1 + 1 + 12) = 3 2 Jawaban C
- 16. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 16 SMAN 12 MAKASSAR 40. Nilai lim 𝑥→0 1−𝑐𝑜𝑠2 𝑥 𝑥 tan 𝑥 = ⋯. A. 3 D. −1 B. 1 E. −3 C. 0 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠2 𝑥 𝑥 tan 𝑥 = lim 𝑥→0 𝑠𝑖𝑛2 𝑥 𝑥 tan 𝑥 = lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 sin 𝑥 tan 𝑥 = 1.1 = 1 Jawaban B 41. Nilai lim 𝑥→0 𝑥 .tan 2𝑥 1−𝑐𝑜𝑠2 𝑥 = ⋯. A. 1 4 D. 2 B. 1 2 E. 4 C. 1 Pembahasan lim 𝑥→0 𝑥 . tan 2𝑥 1 − 𝑐𝑜𝑠2 𝑥 = lim 𝑥→0 𝑥 . tan 2𝑥 𝑠𝑖𝑛2 𝑥 = lim 𝑥→0 𝑥 sin 𝑥 . lim 𝑥→0 tan 2𝑥 sin 𝑥 = 1.2 = 2 Jawaban C 42. Nilai lim 𝑥→0 1−𝑐𝑜𝑠2 𝑥 𝑥2 tan(𝑥+ 𝜋 4 ) = ⋯. A. −1 D. 1 2 √2 B. 0 E. √3 C. 1 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠2 𝑥 𝑥2 tan (𝑥 + 𝜋 4 ) = lim 𝑥→0 𝑠𝑖𝑛2 𝑥 𝑥2 tan (𝑥 + 𝜋 4 ) = lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 1 tan (𝑥 + 𝜋 4 ) = 1.1. 1 tan (0 + 𝜋 4 ) = 1. 1 1 = 1 Jawaban C 43. Nilai lim 𝑥→0 1−𝑐𝑜𝑠2 𝑥 𝑥2 cot(𝑥− 𝜋 3 ) = ⋯. A. 1 D.−√2 B. 0 E.−√3 C. − 1 3 √3
- 17. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 17 SMAN 12 MAKASSAR Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠2 𝑥 𝑥2 cot (𝑥 − 𝜋 3 ) = lim 𝑥→0 𝑠𝑖𝑛2 𝑥 𝑥2 cot (𝑥 − 𝜋 3 ) = lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 1 cot (𝑥 − 𝜋 3 ) = 1.1. 1 cot (0 − 𝜋 3 ) = 1 cot (− 𝜋 3 ) = − tan 𝜋 3 = −√3 Jawaban E 44. Nilai lim 𝑥→0 𝑐𝑜𝑠2 𝑥−1 2 sin 2𝑥 tan 𝑥 = ⋯. A. −4 D.− 1 2 B. −2 E.− 1 4 C. −1 Pembahasan lim 𝑥→0 𝑐𝑜𝑠2 𝑥 − 1 2 sin 2𝑥 tan 𝑥 = lim 𝑥→0 −𝑠𝑖𝑛2 𝑥 2 sin 2𝑥 tan 𝑥 = − 1 2 lim 𝑥→0 sin 𝑥 sin 2𝑥 . lim 𝑥→0 sin 𝑥 tan 𝑥 = − 1 2 . 1 2 . 1 Jawaban E 45. Nilai lim 𝑥→1 1−𝑐𝑜𝑠2(𝑥−1) 4𝑥2−8𝑥+4 = ⋯. A. 0 D.1 B. 1 4 E. 2 C. 1 2 Pembahasan lim 𝑥→1 1 − 𝑐𝑜𝑠2(𝑥 − 1) 4𝑥2 − 8𝑥 + 4 = lim 𝑥→1 𝑠𝑖𝑛2(𝑥 − 1) 4(𝑥2 − 2𝑥 + 1) = 1 4 lim 𝑥→1 sin(𝑥 − 1) (𝑥 − 1) . lim 𝑥→1 sin(𝑥 − 1) (𝑥 − 1) = 1 4 . 1.1 = 1 4 Jawaban B
- 18. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 18 SMAN 12 MAKASSAR 46. Nilai lim x→2 x2−4x+4 1−cos2(x−2) =…. A. − 1 4 D. 1 2 B. 0 E. 1 C. 1 4 Pembahasan lim x→2 x2 − 4x + 4 1 − cos2(x − 2) = lim x→2 (𝑥 − 2)2 sin2(x − 2) = lim x→2 (𝑥 − 2) sin(x − 2) . lim x→2 (𝑥 − 2) sin(x − 2) = 1.1 = 1 Jawaban E 47. Nilai lim 𝑥→0 2𝑥 tan 3𝑥 1−𝑐𝑜𝑠2 𝑥 = ⋯. A. 0 D.6 B. 2 E.12 C. 3 Pembahasan lim 𝑥→0 2𝑥 . tan 3𝑥 1 − 𝑐𝑜𝑠2 𝑥 = lim 𝑥→0 2𝑥 . tan 3𝑥 𝑠𝑖𝑛2 𝑥 = lim 𝑥→0 2𝑥 sin 𝑥 . lim 𝑥→0 tan 3𝑥 sin 𝑥 = 2.3 = 6 Jawaban D 48. Nilai lim 𝑥→0 1−𝑐𝑜𝑠 𝑥 sin 𝑥 = ⋯. A. 0 D.1 B. 1 4 E. 2 C. 1 2 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠 𝑥 sin 𝑥 = lim 𝑥→0 2 𝑠𝑖𝑛2 1 2 𝑥 sin 𝑥 = 2. lim 𝑥→0 sin 1 2 𝑥 sin 𝑥 . lim 𝑥→0 sin 1 2 𝑥 = 2. 1 2 . sin 1 2 . 0 = 0 Jawaban A 49. Nilai lim 𝑥→0 sin23x 1−cos 𝑥 =…. A. 3 D. 12 B. 6 E. 18 C. 9 Pembahasan lim 𝑥→0 sin2 3x 1 − cos 𝑥 = lim 𝑥→0 sin2 3x 2 𝑠𝑖𝑛2 1 2 𝑥 = 1 2 lim 𝑥→0 sin 3𝑥 sin 1 2 𝑥 . lim 𝑥→0 sin 3𝑥 sin 1 2 𝑥
- 19. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 19 SMAN 12 MAKASSAR = 1 2 . 3 1 2 . 3 1 2 = 9 1 2 = 18 Jawaban E 50. Nilai lim 𝜃→0 1−cos 𝜃 𝜃2 = ⋯. A. − 1 4 D. 1 4 B. − 1 2 E. 1 2 C. 0 Pembahasan lim 𝜃→0 1 − cos 𝜃 𝜃2 = lim 𝜃→0 2 𝑠𝑖𝑛2 1 2 𝜃 𝜃2 = 2 lim 𝜃→0 𝑠𝑖𝑛2 1 2 𝜃 𝜃2 = 2 lim 𝜃→0 𝑠𝑖𝑛 1 2 𝜃 𝜃 . lim 𝜃→0 𝑠𝑖𝑛 1 2 𝜃 𝜃 = 2. 1 2 . 1 2 = 1 2 Jawaban E 51. Nilai lim 𝑥→0 1−cos8𝑥 4𝑥2 =…. A. 0 D. 4 B. 1 E. 8 C. 2 Pembahasan lim 𝑥→0 1 − cos 8𝑥 4𝑥2 = lim 𝑥→0 2 𝑠𝑖𝑛2 4𝑥 4𝑥2 = 2 4 lim 𝑥→0 𝑠𝑖𝑛2 4𝑥 𝑥2 = 1 2 lim 𝑥→0 sin 4𝑥 𝑥 . lim 𝑥→0 sin 4𝑥 𝑥 = 1 2 . 4.4 = 8 Jawaban E
- 20. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 20 SMAN 12 MAKASSAR 52. Nilai lim 𝑥→0 1−𝑐𝑜𝑠 2𝑥 1−cos4𝑥 =…. A. − 1 2 B. − 1 4 C. 0 D. 1 16 E. 1 4 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠 2𝑥 1 − cos 4𝑥 = lim x→0 2 sin2 x 2 sin22x = lim x→0 sin2 x sin22x = lim 𝑥→0 sin 𝑥 sin 2𝑥 . lim 𝑥→0 sin 𝑥 sin 2𝑥 = 1 2 . 1 2 = 1 4 Jawaban E 53. Nilai lim 𝑥→0 1−𝑐𝑜𝑠 2𝑥 𝑥 𝑡𝑎𝑛𝑥 = ⋯. A. −8 D.2 B. 0 E.4 C. 1 Pembahasan lim 𝑥→0 1 − 𝑐𝑜𝑠 2𝑥 𝑥 𝑡𝑎𝑛𝑥 = lim 𝑥→0 2𝑠𝑖𝑛2 𝑥 𝑥 𝑡𝑎𝑛𝑥 = 2 lim 𝑥→0 sin 𝑥 sin 𝑥 𝑥 𝑡𝑎𝑛𝑥 = 2 lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 sin 𝑥 tan 𝑥 = 2.1.1 = 2 Jawaban D 54. Nilai lim 𝑥→0 (1−cos 4𝑥) sin 𝑥 𝑥2 tan 3𝑥 = ⋯. A. 128 3 D. 8 3 B. 32 3 E. 4 3 C. 16 3 Pembahasan lim 𝑥→0 (1 − cos 4𝑥) sin 𝑥 𝑥2 tan 3𝑥 = lim 𝑥→0 2. 𝑠𝑖𝑛2 2𝑥. sin 𝑥 𝑥2 tan 3𝑥
- 21. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 21 SMAN 12 MAKASSAR = 2 lim 𝑥→0 𝑠𝑖𝑛2 2𝑥. sin 𝑥 𝑥2 tan 3𝑥 = 2 (lim 𝑥→0 sin 2𝑥 𝑥 ) 2 . lim 𝑥→0 sin 𝑥 tan 3𝑥 = 2. 22 . 1 3 = 8 3 Jawaban D 55. Nilai lim 𝑥→0 cos 4𝑥 −1 𝑥 tan 2𝑥 A. 4 D.−2 B. 2 E. −4 C. −1 Pembahasan lim 𝑥→0 cos 4𝑥 − 1 𝑥 tan 2𝑥 = lim 𝑥→0 −2𝑠𝑖𝑛2 2𝑥 𝑥 tan 2𝑥 = −2 lim 𝑥→0 𝑠𝑖𝑛2 2𝑥 𝑥 tan 2𝑥 = −2. lim 𝑥→0 sin 2𝑥 𝑥 . lim 𝑥→0 sin 2𝑥 tan 2𝑥 = −2.2. 2 2 = −4 Jawaban E 56. Nilai lim 𝑥→0 cos 6𝑥 −1 𝑥 sin 1 2 𝑥 A. 36 D.−9 B. 9 E. −36 C. 0 Pembahasan lim 𝑥→0 cos 6𝑥 − 1 𝑥 sin 1 2 𝑥 = lim 𝑥→0 −2𝑠𝑖𝑛2 3𝑥 𝑥 sin 1 2 𝑥 = −2 lim 𝑥→0 𝑠𝑖𝑛2 3𝑥 𝑥 sin 1 2 𝑥 = −2. lim 𝑥→0 sin 3𝑥 𝑥 . lim 𝑥→0 sin 3𝑥 sin 1 2 𝑥 = −2.3. 3 1 2 = −36 Jawaban E
- 22. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 22 SMAN 12 MAKASSAR 57. Nilai lim 𝑥→0 4x cos6𝑥 −4𝑥 (2𝑥)2.sin5𝑥 = A. − 18 5 D. 2 B. − 5 18 E. 18 5 C. 5 18 Pembahasan lim 𝑥→0 4x cos 6𝑥 − 4𝑥 (2𝑥)2. sin 5𝑥 = lim 𝑥→0 4𝑥(cos 6𝑥 − 1) (2𝑥)2. sin 5𝑥 = lim 𝑥→0 4𝑥(−2 𝑠𝑖𝑛2 3𝑥) (2𝑥)2. sin 5𝑥 = lim 𝑥→0 −8𝑥. sin 3𝑥 sin 3𝑥 (2𝑥)2. sin 5𝑥 = lim 𝑥→0 −8𝑥 sin 5𝑥 . lim 𝑥→0 sin 3𝑥 2𝑥 . lim 𝑥→0 sin 3𝑥 2𝑥 = −8 5 . 3 2 . 3 2 = − 18 5 Jawaban A 58. Jika diketahui 𝑚 = lim 𝑥→0 cos 𝑥−1 cos2𝑥−1 dan 𝑛 = lim 𝑥→2 [ 1 𝑥−2 − 4 𝑥2−4 ], maka 𝑚 + 𝑛 =…. A. −1 D. 1 2 B. − 1 2 C. 1 C. 0 Pembahasan 𝑚 = lim 𝑥→0 cos 𝑥 − 1 cos 2𝑥 − 1 𝑚 = lim 𝑥→0 −2 𝑠𝑖𝑛2 1 2 𝑥 −2 𝑠𝑖𝑛2 𝑥 𝑚 = lim 𝑥→0 𝑠𝑖𝑛2 1 2 𝑥 𝑠𝑖𝑛2 𝑥 𝑚 = (lim 𝑥→0 𝑠𝑖𝑛 1 2 𝑥 𝑠𝑖𝑛𝑥 ) 2 𝑚 = ( 1 2 ) 2 𝑚 = 1 4 𝑛 = lim 𝑥→2 [ 1 𝑥 − 2 − 4 𝑥2 − 4 ] 𝑛 = lim 𝑥→2 [ 𝑥 + 2 𝑥2 − 4 − 4 𝑥2 − 4 ] 𝑛 = lim 𝑥→2 [ 𝑥 + 2 − 4 𝑥2 − 4 ] 𝑛 = lim 𝑥→2 [ 𝑥 − 2 𝑥2 − 4 ] 𝑛 = lim 𝑥→2 𝑥 − 2 (𝑥 − 2)(𝑥 + 2) 𝑛 = lim 𝑥→2 1 (𝑥 + 2) 𝑛 = 1 2 + 2 𝑛 = 1 4 Nilai 𝑚 + 𝑛 = 1 4 + 1 4 = 1 2 Jawaban D
- 23. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 23 SMAN 12 MAKASSAR 59. Nilai dari lim 𝑥→−2 (𝑥2−4) tan(𝑥+2) sin2(x+2) =…. A. −4 D. 4 B. −3 E. 5 C. 0 Pembahasan lim 𝑥→−2 (𝑥2 − 4) tan(𝑥 + 2) sin2(x + 2) = lim 𝑥→−2 (𝑥 − 2)(𝑥 + 2) tan(𝑥 + 2) sin2(x + 2) = lim 𝑥→−2 (𝑥 − 2) lim 𝑥→−2 (𝑥 + 2) sin(x + 2) . lim 𝑥→−2 tan(𝑥 + 2) sin(x + 2) = (−2 − 2).1.1 = −4 Jawaban A 60. Nilai lim 𝑥→3 𝑥 tan(2𝑥−6) sin(𝑥−3) = ⋯. A. 6 D. 1 B. 3 E.0 C. 2 Pembahasan lim 𝑥→3 𝑥 tan(2𝑥 − 6) sin(𝑥 − 3) = lim 𝑥→3 𝑥. lim 𝑥→3 tan(2𝑥 − 6) sin(𝑥 − 3) = lim 𝑥→3 𝑥. lim 𝑥→3 tan 2(𝑥 − 3) sin(𝑥 − 3) = 3.2 =6 Jawaban A 61. Nilai lim 𝑥→0 cos 𝑥−cos5𝑥 1−cos 4𝑥 = ⋯. A. 1 3 D.2 B. 1 2 E.3 C. 3 2 Pembahasan lim 𝑥→0 cos 𝑥 − cos 5𝑥 1 − cos 4𝑥 = lim 𝑥→0 −2 sin 3𝑥 . sin(−2𝑥) 2 𝑠𝑖𝑛22𝑥 = 2 2 lim 𝑥→0 sin 3𝑥 . sin 2𝑥 𝑠𝑖𝑛22𝑥 = lim 𝑥→0 sin 3𝑥 . sin 2𝑥 𝑠𝑖𝑛22𝑥 = lim 𝑥→0 sin 3𝑥 . sin 2𝑥 sin 2𝑥 . sin 2𝑥 = lim 𝑥→0 sin 3𝑥 sin 2𝑥 Rumus cos 𝐴 − cos 𝐵 = −2𝑠𝑖𝑛 1 2 (𝐴 + 𝐵) sin 1 2 (𝐴 − 𝐵)
- 24. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 24 SMAN 12 MAKASSAR = 3 2 Jawaban C 62. lim 𝑥→0 𝑥.tan 5𝑥 cos2𝑥−cos 7𝑥 = ⋯. A. 2 9 D. − 1 9 B. 1 9 E. − 2 9 C. 0 Pembahasan lim 𝑥→0 𝑥. tan 5𝑥 cos 2𝑥 − cos 7𝑥 = lim 𝑥→0 𝑥. tan 5𝑥 −2 sin 9 2 𝑥 sin (− 5 2 ) 𝑥 = 1 2 lim 𝑥→0 𝑥. tan 5𝑥 sin 9 2 𝑥 sin 5 2 𝑥 = 1 2 lim 𝑥→0 𝑥 sin 9 2 𝑥 . lim 𝑥→0 tan 5𝑥 sin 5 2 𝑥 = 1 2 . 1 9 2 . 5 5 2 = 2 9 Jawaban A 63. Nilai lim 𝑥→0 1−cos 8𝑥 sin 2𝑥 tan 2𝑥 = ⋯. A. 16 D. 4 B. 12 E. 2 C. 8 Pembahasan lim 𝑥→0 1 − cos 8𝑥 sin 2𝑥 tan 2𝑥 = lim 𝑥→0 2 𝑠𝑖𝑛2 4𝑥 sin 2𝑥 tan 2𝑥 = lim 𝑥→0 sin 4𝑥 . sin 4𝑥 sin 2𝑥 tan 2𝑥 = lim 𝑥→0 sin 4𝑥 sin 2𝑥 . lim 𝑥→0 sin 4𝑥 tan 2𝑥 = 4 2 . 4 2 = 4 Jawaban D 64. Nilai lim 𝑥→0 1−2𝑠𝑖𝑛2 𝑥−𝑐𝑜𝑠32𝑥 5𝑥2 = ⋯. A. 4 25 D. 4 5 B. 2 5 E.1 C. 3 5
- 25. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 25 SMAN 12 MAKASSAR Pembahasan lim 𝑥→0 1 − 2𝑠𝑖𝑛2 𝑥 − 𝑐𝑜𝑠3 2𝑥 5𝑥2 = lim x→0 cos 2x − cos3 2x 5x2 = lim x→0 cos 2x (1 − cos2 2x) 5x2 = lim x→0 cos 2x sin2 2x 5x2 = 1 5 lim x→0 cos 2𝑥. (lim x→0 sin 2𝑥 𝑥 ) 2 = 1 5 . cos 2.0 . (2)2 = 1 5 . 1.4 = 4 5 karena 𝟏 − 𝟐𝒔𝒊𝒏 𝟐 𝒙 = 𝐜𝐨𝐬 𝟐𝒙 Jawaban D 65. Nilai lim 𝑥→0 1−cos2x−cos x sin2x 𝑥4 = ⋯. A. −1 D. 1 2 B. 0 E. 1 C. 1 4 Pembahasan lim 𝑥→0 1 − cos2 x − cos x sin2 x 𝑥4 = lim 𝑥→0 sin2 x − cos x sin2 x 𝑥4 = lim 𝑥→0 sin2 x(1 − cos x) 𝑥4 = lim 𝑥→0 sin2 x. 2 𝑠𝑖𝑛2 1 2 x 𝑥4 = 2 lim 𝑥→0 sin2 x. 𝑠𝑖𝑛2 1 2 x 𝑥4 = 2 lim 𝑥→0 sin2 x 𝑥2 . lim 𝑥→0 sin2 1 2 x 𝑥2 = 2 (lim 𝑥→0 sin 𝑥 𝑥 ) 2 . (lim 𝑥→0 sin 1 2 𝑥 𝑥 ) 2 = 2. 12 . ( 1 2 ) 2 =2. 1 4 = 2 4 = 1 2 Jawaban D
- 26. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 26 SMAN 12 MAKASSAR 66. Nilai lim 𝑥→0 5𝑥2−2𝑥 sin 5𝑥−tan 2𝑥 = ⋯. A. − 2 7 D.− 5 3 B. − 2 3 E.− 5 2 C. −1 Pembahasan lim 𝑥→0 5𝑥2 − 2𝑥 sin 5𝑥 − tan 2𝑥 = lim 𝑥→0 5𝑥2 𝑥 − 2𝑥 𝑥 sin 5𝑥 𝑥 − tan 2𝑥 𝑥 = lim 𝑥→0 5𝑥 − 2 sin 5𝑥 𝑥 − tan 2𝑥 𝑥 = lim 𝑥→0 5𝑥 − lim 𝑥→0 2 lim 𝑥→0 sin 5𝑥 𝑥 − tan 2𝑥 𝑥 = 0 − 2 5 − 2 = − 2 3 Jawaban B 67. Nilai lim 𝑥→0 𝑥2+𝑠𝑖𝑛23𝑥 2𝑡𝑎𝑛2 𝑥2 = ⋯. A. 5 D. 1 2 B. 2 E. 2 5 C. 1 Pembahasan lim 𝑥→0 𝑥2 + 𝑠𝑖𝑛2 3𝑥 2𝑡𝑎𝑛2 𝑥2 = 1 2 lim 𝑥→0 𝑥2 + 𝑠𝑖𝑛2 3𝑥 𝑡𝑎𝑛2 𝑥2 = 1 2 . lim 𝑥→0 1 + lim 𝑥→0 𝑠𝑖𝑛2 3𝑥 𝑥2 lim 𝑥→0 𝑡𝑎𝑛2 𝑥2 𝑥2 = 1 2 . 1 + 9 1 = 10 2 =5 Jawaban A 68. Nilai lim 𝑥→0 2𝑥2+𝑥 sin 𝑥 = ⋯. A. −1 D. 2 B. 0 E. 3 C. 1 Pembahasan lim 𝑥→0 2𝑥2 + 𝑥 sin 𝑥 = lim 𝑥→0 𝑥(2𝑥 + 1) sin 𝑥
- 27. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 27 SMAN 12 MAKASSAR = lim 𝑥→0 𝑥 sin 𝑥 . lim 𝑥→0 (2𝑥 + 1) = 1. (2.0 + 1) = 1.1 =1 Jawaban C 69. Nilai lim 𝑥→0 𝑥2+2𝑥 tan 𝑥 = ⋯. A. 2 D. 1 4 B. 1 E.0 C. 1 2 Pembahasan lim 𝑥→0 𝑥2 + 2𝑥 tan 𝑥 = lim 𝑥→0 𝑥(𝑥 + 2) tan 𝑥 = lim 𝑥→0 𝑥 tan 𝑥 . lim 𝑥→0 (𝑥 + 2) = 1. (0 + 2) = 1.2 =2 Jawaban A 70. Nilai lim 𝑥→0 √1+tan 𝑥−√1+sin 𝑥 𝑥3 = ⋯. A. −1 D. 1 4 B. − 1 4 E.1 C. 0 Pembahasan lim 𝑥→0 √1 + tan 𝑥 − √1 + sin 𝑥 𝑥3 = lim 𝑥→0 √1 + tan 𝑥 − √1 + sin 𝑥 𝑥3 × √1 + tan 𝑥 + √1 + sin 𝑥 √1 + tan 𝑥 + √1 + sin 𝑥 = lim 𝑥→0 (1 + tan 𝑥) − (1 + sin 𝑥) 𝑥3 (√1 + tan 𝑥 + √1 + sin 𝑥) = lim 𝑥→0 tan 𝑥 − sin 𝑥 𝑥3 (√1 + tan 𝑥 + √1 + sin 𝑥) = lim 𝑥→0 sin 𝑥 cos 𝑥 − sin 𝑥 𝑥3 (√1 + tan 𝑥 + √1 + sin 𝑥) = lim 𝑥→0 sin 𝑥 − sin 𝑥 . cos 𝑥 cos 𝑥 𝑥3 (√1 + tan 𝑥 + √1 + sin 𝑥) = lim 𝑥→0 sin 𝑥 (1 − cos 𝑥) 𝑥3 cos 𝑥 (√1 + tan 𝑥 + √1 + sin 𝑥) = lim 𝑥→0 sin 𝑥 (2. 𝑠𝑖𝑛2 1 2 𝑥) 𝑥3 cos 𝑥 (√1 + tan 𝑥 + √1 + sin 𝑥)
- 28. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 28 SMAN 12 MAKASSAR = 2lim 𝑥→0 sin 𝑥 𝑥 . (lim 𝑥→0 sin 1 2 𝑥 𝑥 ) 2 . lim 𝑥→0 1 cos 𝑥 (√1 + tan 𝑥 + √1 + sin 𝑥) = 2.1. ( 1 2 ) 2 . 1 cos 0 (√1 + tan 0 + √1 + sin 0) = 2. 1 4 . 1 1(√1 + √1) = 2 8 = 1 4 Jawaban D 71. Nilai lim 𝑥→0 √1+sin 𝑥−√1−sin 𝑥 𝑥 = ⋯. A. −1 D. √2 B. − 1 4 E. 1 C. 1 4 √2 Pembahasan lim 𝑥→0 √1 + sin 𝑥 − √1 − sin 𝑥 𝑥 = lim 𝑥→0 √1 + sin 𝑥 − √1 − sin 𝑥 𝑥 × √1 + sin 𝑥 + √1 − sin 𝑥 √1 + sin 𝑥 + √1 − sin 𝑥 = lim 𝑥→0 (1 + sin 𝑥) − (1 − sin 𝑥) 𝑥(√1 + sin 𝑥 + √1 − sin 𝑥) = lim 𝑥→0 2 sin 𝑥 𝑥(√1 + sin 𝑥 + √1 − sin 𝑥) = lim 𝑥→0 2 sin 𝑥 𝑥 . lim 𝑥→0 1 (√1 + sin 𝑥 + √1 − sin 𝑥) = 2. 1 (√1 + sin 0 + √1 − sin 0) = 2. 1 (1 + 1) = 2 2 = 1 Jawaban E 72. Nilai lim 𝑥→0 (1−√cos 𝑥) cot 𝑥 𝑥 = ⋯. A. − 1 2 D. 1 4 B. − 1 4 E. 1 2 C. 0 Pembahasan lim 𝑥→0 (1 − √cos 𝑥) cot 𝑥 𝑥
- 29. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 29 SMAN 12 MAKASSAR = lim 𝑥→0 (1 − √cos 𝑥) cot 𝑥 𝑥 × (1 + √cos 𝑥) (1 + √cos 𝑥) = lim 𝑥→0 (1 − cos 𝑥) cot 𝑥 𝑥(1 + √cos 𝑥) = lim 𝑥→0 2 sin 2 1 2 x 𝑥(1 + √cos 𝑥) tan 𝑥 = lim 𝑥→0 2 sin 1 2 𝑥 𝑥 . lim 𝑥→0 sin 1 2 𝑥 tan 𝑥 . lim 𝑥→0 1 (1 + √cos 𝑥) = 2. 1 2 . 1 2 . 1 1 + √cos 0 = 1 2 . 1 2 = 1 4 Jawaban D 73. Nilai lim 𝑥→3 𝑥2−9 sin(𝑥−3) = ⋯. A. 9 D.3 B. 7 E.1 C. 6 Pembahasan lim 𝑥→3 𝑥2 − 9 sin(𝑥 − 3) = lim 𝑥→3 (𝑥 − 3)(𝑥 + 3) sin(𝑥 − 3) = lim 𝑥→3 (𝑥 − 3) sin(𝑥 − 3) . lim 𝑥→3 (𝑥 + 3) = 1. (3 + 3) = 6 Jawaban C 74. Nilai lim 𝑥→0 (𝑥2−1) sin 6𝑥 𝑥3+3𝑥2+2𝑥 = ⋯. A. −3 D. 1 B. −1 E. 6 C. 0 Pembahasan lim 𝑥→0 (𝑥2 − 1) sin 6𝑥 𝑥3 + 3𝑥2 + 2𝑥 = lim 𝑥→0 (𝑥2 − 1) sin 6𝑥 𝑥(𝑥2 + 3𝑥 + 2) = lim 𝑥→1 (𝑥2 − 1) (𝑥2 + 3𝑥 + 2) . lim 𝑥→1 sin 6𝑥 𝑥 = (02 − 1) (02 + 3.0 + 2) . 6 = −1 2 . 6 = −3 Jawaban A
- 30. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 30 SMAN 12 MAKASSAR 75. Nilai lim 𝑥→1 𝑥3−(𝑎+1)𝑥2+𝑎𝑥 (𝑥2−𝑎)tan(𝑥−1) = ⋯. A. 1 D. 0 B. 1 − 𝑎 E. 2 − 𝑎 C. 𝑎 Pembahasan lim 𝑥→1 𝑥3 − (𝑎 + 1)𝑥2 + 𝑎𝑥 (𝑥2 − 𝑎) tan(𝑥 − 1) = lim 𝑥→1 𝑥(𝑥2 − (𝑎 + 1)𝑥 + 𝑎) (𝑥2 − 𝑎) tan(𝑥 − 1) = lim 𝑥→1 𝑥(𝑥 − 𝑎)(𝑥 − 1) (𝑥2 − 𝑎) tan(𝑥 − 1) = lim 𝑥→1 𝑥(𝑥 − 𝑎) (𝑥2 − 𝑎) . lim 𝑥→1 (𝑥 − 1) tan(𝑥 − 1) = 1(1 − 𝑎) (12 − 𝑎) . 1 = 1 − 𝑎 1 − 𝑎 = 1 Jawaban A 76. Nilai lim 𝑥→1 (𝑥2−1) sin 2(𝑥−1) −2 𝑠𝑖𝑛2(𝑥−1) = ⋯. A. −2 D.− 1 4 B. −1 E.0 C. − 1 2 Pembahasan lim 𝑥→1 (𝑥2 − 1) sin 2(𝑥 − 1) −2 𝑠𝑖𝑛2(𝑥 − 1) = lim 𝑥→1 (𝑥 − 1)(𝑥 + 1) sin 2(𝑥 − 1) −2 𝑠𝑖𝑛2(𝑥 − 1) = − 1 2 lim 𝑥→1 (𝑥 − 1)(𝑥 + 1) sin 2(𝑥 − 1) sin(𝑥 − 1) . sin(𝑥 − 1) = − 1 2 lim 𝑥→1 (𝑥 − 1) sin(𝑥 − 1) . lim 𝑥→1 sin2(𝑥 − 1) sin(𝑥 − 1) . lim 𝑥→1 (𝑥 + 1) =− 1 2 . 1.2. (1 + 1) = −2 Jawaban A 77. Nilai lim 𝑥→0 2 𝑠𝑖𝑛2 𝑥 2 𝑥 sin 𝑥 = ⋯. A. 0 D. 2 B. 1 4 E. 4 C. 1 2
- 31. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 31 SMAN 12 MAKASSAR Pembahasan lim 𝑥→0 2 𝑠𝑖𝑛2 𝑥 2 𝑥 sin 𝑥 = 2 lim 𝑥→0 sin 𝑥 2 . sin 𝑥 2 𝑥 sin 𝑥 = 2 lim 𝑥→0 sin 𝑥 2 𝑥 . lim 𝑥→0 sin 𝑥 2 sin 𝑥 = 2. 1 2 . 1 2 = 1 2 Jawaban C 78. Nilai lim 𝑥→0 sin 4𝑥 1−√1−𝑥 = ⋯. A. 8 D. −6 B. 6 E.−8 C. 4 Pembahasan lim 𝑥→0 sin 4𝑥 1 − √1 − 𝑥 = lim 𝑥→0 sin 4𝑥 1 − √1 − 𝑥 . 1 + √1 − 𝑥 1 + √1 − 𝑥 = lim 𝑥→0 sin 4𝑥(1 + √1 − 𝑥) 1 − (1 − 𝑥) = lim 𝑥→0 sin 4𝑥(1 + √1 − 𝑥) 𝑥 = lim 𝑥→0 sin 4𝑥 𝑥 . lim 𝑥→0 (1 + √1 − 𝑥) = 4. (1 + √1 − 0) = 4(1 + 1) =8 Jawaban A 79. Nilai lim 𝑥→ 𝜋 3 tan(3𝑥−𝜋)cos 2𝑥 sin(3𝑥−𝜋) = ⋯. A. − 1 2 D. 1 2 √3 B. 1 2 E. 3 2 C. 1 2 √2 Pembahasan lim 𝑥→ 𝜋 3 tan(3𝑥 − 𝜋) cos 2𝑥 sin(3𝑥 − 𝜋) = lim 𝑥→ 𝜋 3 tan(3𝑥 − 𝜋) sin(3𝑥 − 𝜋) . lim 𝑥→ 𝜋 3 cos 2𝑥 = 1. cos (2. 𝜋 3 ) = cos ( 2𝜋 3 ) = − 1 2 Jawaban A
- 32. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 32 SMAN 12 MAKASSAR 80. Nilai dari lim 𝑥→2 (𝑥−2) cos(𝜋𝑥−2𝜋) tan(2𝜋𝑥−4𝜋) =…. A. 2𝜋 D. 1 𝜋 B. 𝜋 E. 1 2𝜋 C. 0 Pembahasan lim 𝑥→2 (𝑥 − 2) cos(𝜋𝑥 − 2𝜋) tan(2𝜋𝑥 − 4𝜋) = lim 𝑥→2 (𝑥 − 2) cos 𝜋(𝑥 − 2) tan2𝜋(𝑥 − 2) = lim 𝑥→2 (𝑥 − 2) tan2𝜋(𝑥 − 2) . lim 𝑥→2 cos 𝜋(𝑥 − 2) = 1 2𝜋 . cos 𝜋(2 − 2) = 1 2𝜋 . cos 0 = 1 2𝜋 . 1 = 1 2𝜋 Jawaban E 81. Nilai lim 𝑥→−3 𝑥2+6𝑥+9 2−2 cos(2𝑥+6) = ⋯. A. 3 D. 1 3 B. 1 E. 1 4 C. 1 2 Pembahasan lim 𝑥→−3 𝑥2 + 6𝑥 + 9 2 − 2 cos(2𝑥 + 6) = lim 𝑥→−3 (𝑥 + 3)2 2(1 − cos(2𝑥 + 6)) = 1 2 lim 𝑥→−3 (𝑥 + 3)2 (1 − cos 2(𝑥 + 3)) = 1 2 lim 𝑥→−3 (𝑥 + 3)2 2 sin2(𝑥 + 3) = 1 4 lim 𝑥→−3 (𝑥 + 3)2 sin2(𝑥 + 3) = 1 4 { lim 𝑥→−3 (𝑥 + 3) sin(𝑥 + 3) } 2 = 1 4 . 12 = 1 4 Jawaban E 82. Nilai lim 𝑥→−2 2−2 cos(𝑥+2) 𝑥2+4𝑥+4 =…. A. 4 D. 1 B. 2 E. 1 2 C. 1 4
- 33. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 33 SMAN 12 MAKASSAR Pembahasan lim 𝑥→−2 2 − 2 cos(𝑥 + 2) 𝑥2 + 4𝑥 + 4 = lim 𝑥→−2 2(1 − cos(𝑥 + 2)) (𝑥 + 2)2 = 2 lim 𝑥→−2 2. 𝑠𝑖𝑛2 1 2 (𝑥 + 2) (𝑥 + 2)2 = 4 . { lim 𝑥→−2 sin 1 2 (𝑥 + 2) (𝑥 + 2) } 2 = 4 . { 1 2 } 2 = 4. 1 4 = 1 Jawaban D 83. Nilai lim 𝑥→1 tan(𝑥−1) sin(1−√ 𝑥) 𝑥2−2𝑥+1 = ⋯. A. −1 D. 1 2 B. − 1 2 E. 1 C. 0 Pembahasan lim 𝑥→1 tan(𝑥 − 1) sin(1 − √ 𝑥) 𝑥2 − 2𝑥 + 1 = lim 𝑥→1 tan(𝑥 − 1) sin(1 − √ 𝑥) (𝑥 − 1)(𝑥 − 1) = lim 𝑥→1 tan(𝑥 − 1) sin(1 − √ 𝑥) (𝑥 − 1)(√ 𝑥 − 1)(√ 𝑥 + 1) = lim 𝑥→1 tan(𝑥 − 1) (𝑥 − 1) . lim 𝑥→1 sin(1 − √ 𝑥) (√ 𝑥 − 1) . lim 𝑥→1 1 (√ 𝑥 + 1) = 1. (−1). 1 (√1 + 1) = −1 2 Jawaban B 84. lim 𝑥→1 (𝑥2+𝑥−2)sin(𝑥−1) 𝑥2−2𝑥+1 = ⋯. A. 4 D. − 1 4 B. 3 E. − 1 2 C. 0 Pembahasan lim 𝑥→1 (𝑥2 + 𝑥 − 2)sin(𝑥 − 1) 𝑥2 − 2𝑥 + 1 = lim 𝑥→1 (𝑥 + 2)(𝑥 − 1) sin(𝑥 − 1) (𝑥 − 1)(𝑥 − 1) = lim 𝑥→1 (𝑥 + 2) . = lim 𝑥→1 sin(𝑥 − 1) (𝑥 − 1)
- 34. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 34 SMAN 12 MAKASSAR = (1 + 2). 1 = 3 Jawaban B 85. Nilai lim 𝑥→4 (𝑥+2)tan(𝑥−4) 2𝑥2−7𝑥−4 = ⋯. A. 0 D. 3 2 B. 2 3 E. 2 C. 1 Pembahasan lim 𝑥→4 (𝑥 + 2) tan(𝑥 − 4) 2𝑥2 − 7𝑥 − 4 = lim 𝑥→4 (𝑥 + 2) tan(𝑥 − 4) (2𝑥 + 1)(𝑥 − 4) = lim 𝑥→4 (𝑥 + 2) (2𝑥 + 1) . lim 𝑥→4 tan(𝑥 − 4) (𝑥 − 4) = (4 + 2) (2.4 + 1) .1 = 6 9 = 2 3 Jawaban B 86. Nilai dari ekspresi lim 𝑥→2 (2𝑥+1)tan(𝑥−2) (𝑥2−4) sama dengan …. A. 1,25 D. 2,50 B. 1,50 E. 5,00 C. 2,00 Pembahasan lim 𝑥→2 (2𝑥 + 1) tan(𝑥 − 2) (𝑥2 − 4) = lim 𝑥→2 (2𝑥 + 1) tan(𝑥 − 2) (𝑥 + 2)(𝑥 − 2) = lim 𝑥→2 (2𝑥 + 1) (𝑥 + 2) . lim 𝑥→2 tan(𝑥 − 2) (𝑥 − 2) = (2.2 + 1) (2 + 2) = 5 4 =1,25 Jawaban A 87. Nilai lim 𝑥→1 (3𝑥+1)sin(𝑥−1) 𝑥2+2𝑥−3 = ⋯. A. 4 D. 1 B. 3 E. 0 C. 2 Pembahasan lim 𝑥→1 (3𝑥 + 1) sin(𝑥 − 1) 𝑥2 + 2𝑥 − 3 = lim 𝑥→1 (3𝑥 + 1) sin(𝑥 − 1) (𝑥 + 3)(𝑥 − 1)
- 35. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 35 SMAN 12 MAKASSAR = lim 𝑥→1 (3𝑥 + 1) (𝑥 + 3) . lim 𝑥→1 sin(𝑥 − 1) (𝑥 − 1) = (3.1 + 1) (1 + 3) . 1 = 4 4 = 1 Jawaban D 88. Nilai lim 𝑡→2 (𝑡2−5𝑡+6) sin(𝑡−2) (𝑡2−𝑡−2)2 = ⋯. A. 1 3 D.− 1 9 B. 1 9 E.− 1 3 C. 0 Pembahasan lim 𝑡→2 (𝑡2 − 5𝑡 + 6) sin(𝑡 − 2) (𝑡2 − 𝑡 − 2)2 = lim 𝑡→2 (𝑡 − 2)(𝑡 − 3) sin(𝑡 − 2) ((𝑡 − 2)(𝑡 + 1)) 2 = lim 𝑡→2 (𝑡 − 2)(𝑡 − 3) sin(𝑡 − 2) (𝑡 − 2)2(𝑡 + 1)2 = lim 𝑡→2 (𝑡 − 3) sin(𝑡 − 2) (𝑡 + 1)2(𝑡 − 2) = lim 𝑡→2 (𝑡 − 3) (𝑡 + 1)2 . lim 𝑡→2 sin(𝑡 − 2) (𝑡 − 2) = (2 − 3) (2 + 1)2 . 1 = − 1 9 Jawaban D 89. Nilai lim 𝑥→1 1− 1 𝑥 sin 𝜋(𝑥−1) =…. A. 2 𝜋 D. − 1 𝜋 B. 1 𝜋 E. − 2 𝜋 C. 0 Pembahasan lim 𝑥→1 1 − 1 𝑥 sin 𝜋(𝑥 − 1) = lim 𝑥→1 𝑥 − 1 𝑥 sin 𝜋(𝑥 − 1) = lim 𝑥→1 (𝑥 − 1) x. sin 𝜋(𝑥 − 1) = lim 𝑥→1 1 𝑥 . lim 𝑥→1 (𝑥 − 1) sin 𝜋(𝑥 − 1) = 1 1 . 1 𝜋 = 1 𝜋 Jawaban B
- 36. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 36 SMAN 12 MAKASSAR 90. Nilai lim 𝑥→ 𝜋 sin(𝑥−𝜋) 2(𝑥−𝜋)+tan(𝑥−𝜋) = ⋯. A. − 1 2 D. 1 3 B. − 1 4 E. 2 5 C. 1 4 Pembahasan Misalkan 𝑥 − 𝜋 = 𝑦 Jika 𝑥 → 𝜋 maka 𝑦 → 𝜋 − 𝜋 = 0 lim 𝑥→ 𝜋 sin(𝑥 − 𝜋) 2(𝑥 − 𝜋) + tan(𝑥 − 𝜋) = lim 𝑦→ 0 sin 𝑦 2𝑦 + tan 𝑦 = lim 𝑦→ 0 sin 𝑦 𝑦 2𝑦 𝑦 + tan 𝑦 𝑦 = lim 𝑦→ 0 sin 𝑦 𝑦 lim 𝑦→ 0 2 + lim 𝑦→ 0 tan 𝑦 𝑦 = 1 2 + 1 = 1 3 Jawaban D 91. Nilai lim 𝑥→ 𝜋 3 sin(𝑥− 𝜋 3 )+sin 5(𝑥− 𝜋 3 ) 6(𝑥− 𝜋 3 ) = ⋯. A. 1 D.3 B. 2 E. 7 2 C. 5 2 Pembahasan Misalkan 𝑥 − 𝜋 3 = 𝑦 Jika 𝑥 → 𝜋 3 maka 𝑦 → 𝜋 3 − 𝜋 3 = 0 lim 𝑥→ 𝜋 3 sin (𝑥 − 𝜋 3 ) + sin 5 (𝑥 − 𝜋 3 ) 6 (𝑥 − 𝜋 3 ) = lim 𝑦→0 sin 𝑦 + sin 5𝑦 6𝑦 = lim 𝑦→0 sin 𝑦 6𝑦 + lim 𝑦→0 sin 5𝑦 6𝑦 = 1 6 + 5 6 = 1 Jawaban A
- 37. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 37 SMAN 12 MAKASSAR 92. Nilai lim 𝑥→𝑎 𝑥−𝑎 sin(𝑥−𝑎)−2𝑥+2𝑎 = ⋯. A. 6 D. −1 B. 3 E. −3 C. 1 Pembahasan Misalkan 𝑥 − 𝑎 = 𝑦 Jika 𝑥 → 𝑎 maka 𝑦 → 𝑎 − 𝑎 = 0 lim 𝑥→𝑎 𝑥 − 𝑎 sin(𝑥 − 𝑎) − 2𝑥 + 2𝑎 = lim 𝑥→𝑎 𝑥 − 𝑎 sin(𝑥 − 𝑎) − 2(𝑥 − 𝑎) = lim 𝑦→0 𝑦 sin 𝑦 − 2𝑦 = lim 𝑦→0 𝑦 𝑦 sin 𝑦 𝑦 − 2𝑦 𝑦 = lim 𝑦→0 1 lim 𝑦→0 sin 𝑦 𝑦 − lim 𝑦→0 2 = 1 1 − 2 = −1 Jawaban D 93. Jika diketahui lim 𝑥→0 tan 𝑥 𝑥 = 1, maka lim 𝑥→𝑎 𝑥−𝑎 tan(𝑥−𝑎)+3𝑥−3𝑎 =…. A. 0 D. 1 2 B. 1 4 E. 1 C. 1 3 Pembahasan Misalkan 𝑥 − 𝑎 = 𝑦 Jika 𝑥 → 𝑎 maka 𝑦 → 𝑎 − 𝑎 = 0 lim 𝑥→𝑎 𝑥 − 𝑎 tan(𝑥 − 𝑎) + 3𝑥 − 3𝑎 = lim 𝑥→𝑎 𝑥 − 𝑎 tan(𝑥 − 𝑎) + 3(𝑥 − 𝑎) = lim 𝑦→0 𝑦 tan 𝑦 + 3𝑦 = lim 𝑦→0 𝑦 𝑦 tan 𝑦 𝑦 + 3𝑦 𝑦 = lim 𝑦→0 1 lim 𝑦→0 tan 𝑦 𝑦 + lim 𝑦→0 3 = 1 1 + 3 = 1 4 Jawaban B
- 38. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 38 SMAN 12 MAKASSAR 94. Nilai lim 𝑥→0 1−cos 𝑥 cos 3𝑥−cos 𝑥 = ⋯. A. − 1 8 D. 1 4 B. − 1 4 E. 1 8 C. − 1 2 Pembahasan lim 𝑥→0 1 − cos 𝑥 cos 3𝑥 − cos 𝑥 = lim 𝑥→0 2 𝑠𝑖𝑛2 1 2 𝑥 −2 sin 2𝑥 . 𝑠𝑖𝑛𝑥 = lim 𝑥→0 sin 1 2 𝑥 . sin 1 2 𝑥 − sin 2𝑥 . 𝑠𝑖𝑛𝑥 = lim 𝑥→0 sin 1 2 𝑥 − sin 2𝑥 . lim 𝑥→0 sin 1 2 𝑥 sin 𝑥 = − 1 2 2 . 1 2 = − 1 8 Jawaban A 95. Nilai lim 𝑥→0 cos 𝑥−cos 5𝑥 12𝑥 tan 2𝑥 = ⋯. A. 1 6 D.− 1 6 B. 1 2 E. − 1 12 C. − 1 2 Pembahasan lim 𝑥→0 cos 𝑥 − cos 5𝑥 12𝑥 tan 2𝑥 = lim 𝑥→0 −2 sin 3𝑥 sin(−2𝑥) 12𝑥 tan 2𝑥 = 2 12 lim 𝑥→0 sin 3𝑥 sin 2𝑥 𝑥 tan 2𝑥 = 1 6 lim 𝑥→0 sin 3𝑥 𝑥 . lim 𝑥→0 sin 2𝑥 tan 2𝑥 = 1 6 . 3. 2 2 = 1 2 Jawaban B 96. Jika diketahui lim 𝑥→0 sin 𝑥 𝑥 = 1, maka lim 𝑥→0 cos 𝑥−cos 2𝑥 𝑥2 =…. A. 1 2 D. 3 2 B. 2 3 E. 2 C. 1
- 39. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 39 SMAN 12 MAKASSAR Pembahasan lim 𝑥→0 cos 𝑥 − cos 2𝑥 𝑥2 = lim 𝑥→0 2 sin 3 2 𝑥 sin 1 2 𝑥 𝑥2 = 2. lim 𝑥→0 sin 3 2 𝑥 𝑥 . lim 𝑥→0 sin 1 2 𝑥 𝑥 = 2. 3 2 . 1 2 = 3 2 Jawaban D 97. lim 𝑎→0 cos 𝑚𝛼−cos 𝑛𝛼 𝛼2 =…. A. 𝑚−𝑛 2 D. 𝑚+𝑛 2 B. 𝑚2−𝑛2 2 E. 𝑛2−𝑚2 2 C. 𝑚2+𝑛2 2 Pembahasan lim 𝑎→0 cos 𝑚𝛼 − cos 𝑛𝛼 𝛼2 = lim 𝑎→0 −2 sin (𝑚𝛼 + 𝑛𝛼) 2 . sin (𝑚𝛼 − 𝑛𝛼) 2 𝛼2 = lim 𝑎→0 −2 sin (𝑚𝛼 + 𝑛𝛼) 2 𝛼 . lim 𝑎→0 sin (𝑚𝛼 − 𝑛𝛼) 2 𝛼 = lim 𝑎→0 −2 sin 𝛼(𝑚 + 𝑛) 2 𝛼 . lim 𝑎→0 sin 𝛼(𝑚 − 𝑛) 2 𝛼 = −2. (𝑚+𝑛) 2 . (𝑚−𝑛) 2 = (𝑚 + 𝑛)(𝑚 − 𝑛) 2 = 𝑚2 − 𝑛2 2 Jawaban B 98. Nilai dari xx xx x cos sin5sin 0 lim … A. 0 B. 1 C. 3 D. 4 E. 5 Pembahasan lim 𝑥→0 sin 5𝑥 − sin 𝑥 𝑥 cos 𝑥 = lim 𝑥→0 2 cos 3𝑥 sin 2𝑥 𝑥 cos 𝑥
- 40. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 40 SMAN 12 MAKASSAR = lim 𝑥→0 2 cos 3𝑥 cos 𝑥 . lim 𝑥→0 sin 2𝑥 𝑥 = 2 cos 3.0 cos 0 . 2 = 2.1 1 . 2 = 4 Jawaban D 99. Nilai lim 𝑥→0 sin 3𝑥−sin3𝑥 cos2𝑥 2𝑥3 = ⋯. A. 4 D.1 B. 3 E. 1 3 C. 2 Pembahasan lim 𝑥→0 sin 3𝑥 − sin 3𝑥 cos 2𝑥 2𝑥3 = lim 𝑥→0 sin 3𝑥(1 − cos 2𝑥) 2𝑥3 = lim 𝑥→0 sin 3𝑥. 2 𝑠𝑖𝑛2 𝑥 2𝑥3 = lim 𝑥→0 sin 3𝑥. 𝑠𝑖𝑛2 𝑥 𝑥3 = lim 𝑥→0 sin 3𝑥 𝑥 . lim 𝑥→0 sin 𝑥 𝑥 . lim 𝑥→0 sin 𝑥 𝑥 = 3.1.1 = 3 Jawaban B 100. Nilai lim 𝜃→0 tan 𝜃−sin 𝜃 𝜃3 =…. A. 1 4 D. 2 B. 1 2 E. 3 C. 1 Pembahasan lim 𝜃→0 tan 𝜃 − sin 𝜃 𝜃3 = lim 𝜃→0 sin 𝜃 cos 𝜃 − sin 𝜃 𝜃3 = lim 𝜃→0 sin 𝜃 ( 1 cos 𝜃 − 1) 𝜃3 = lim 𝜃→0 sin 𝜃 ( 1 − cos 𝜃 cos 𝜃 ) 𝜃3 = lim 𝜃→0 sin 𝜃 (2 𝑠𝑖𝑛2 1 2 𝜃) cos 𝜃 . 𝜃3
- 41. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 41 SMAN 12 MAKASSAR = 2lim 𝜃→0 tan 𝜃 ( 𝑠𝑖𝑛2 1 2 𝜃) 𝜃3 = 2lim 𝜃→0 tan 𝜃 𝜃 . (lim 𝜃→0 sin 1 2 𝜃 𝜃 ) 2 = 2.1. ( 1 2 ) 2 =2. 1 4 = 1 2 Jawaban B 101. Jika lim 𝑥→0 𝑥3 tan 𝑥−sin 𝑥 = 𝐴 − 2, maka nilai dari (𝐴 + 2) adalah …. A. −2 D.4 B. 0 E. 6 C. 2 Pembahasan lim 𝑥→0 𝑥3 tan 𝑥 − sin 𝑥 = lim 𝑥→0 𝑥3 tan 𝑥 (1 − 𝑐𝑜𝑠𝑥) = lim 𝑥→0 𝑥3 tan 𝑥 . 2 𝑠𝑖𝑛2 1 2 𝑥 = 1 2 lim 𝑥→0 𝑥 tan 𝑥 . lim 𝑥→0 𝑥2 𝑠𝑖𝑛2 1 2 𝑥 = 1 2 lim 𝑥→0 𝑥 tan 𝑥 . (lim 𝑥→0 𝑥 𝑠𝑖𝑛 1 2 𝑥 ) 2 = 1 2 . 1. 1 1 4 = 2 Nilai dari 𝐴 − 2 = 2 sehingga A = 4 Jadi A+2 = 4 + 2 = 6 Jawaban E 102.Nilai lim 𝑥→2 1−𝑐𝑜𝑠2(𝑥−2) 3𝑥2−12𝑥+12 = ⋯. A. 1 3 D.1 B. 1 2 E.2 C. 0 Pembahasan lim 𝑥→2 1 − 𝑐𝑜𝑠2(𝑥 − 2) 3𝑥2 − 12𝑥 + 12 = lim 𝑥→2 sin2(𝑥 − 2) 3(𝑥2 − 4𝑥 + 4) = lim 𝑥→2 sin2(𝑥 − 2) 3(𝑥 − 2)2
- 42. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 42 SMAN 12 MAKASSAR = lim 𝑥→2 sin(𝑥 − 2) 3(𝑥 − 2) . lim 𝑥→2 sin(𝑥 − 2) (𝑥 − 2) = 1 3 . 1 = 1 3 Jawaban A 103. Nilai lim 𝑥→ 𝜋 4 (𝑥− 𝜋 4 ) sin(3𝑥− 3𝜋 4 ) 2(1−sin 2𝑥) = ⋯. A. 3 4 D.− 3 4 B. 1 4 E. − 3 2 C. 0 Pembahasan Misalkan 𝑥 − 𝜋 4 = 𝑦 ⇔ 𝑥 = 𝑦 + 𝜋 4 sehingga 2𝑥 = 2𝑦 + 𝜋 2 Jika 𝑥 → 𝜋 4 maka 𝑦 → 0 lim 𝑥→ 𝜋 4 (𝑥 − 𝜋 4 ) sin (3𝑥 − 3𝜋 4 ) 2(1 − sin 2𝑥) = lim 𝑥→ 𝜋 4 (𝑥 − 𝜋 4 ) sin 3 (𝑥 − 𝜋 4 ) 2(1 − sin 2𝑥) = lim 𝑦→0 𝑦 sin 3𝑦 2 (1 − sin (2𝑦 + 𝜋 2 )) = lim 𝑦→0 𝑦 sin 3𝑦 2. (1 − 𝑐𝑜𝑠2𝑦) = 1 2 lim 𝑦→0 𝑦 sin 3𝑦 2. sin2y = 1 4 . lim 𝑦→0 𝑦 sin 𝑦 . lim 𝑦→0 sin 3𝑦 sin 𝑦 = 1 4 . 1.3 = 3 4 Jawaban A 104. Nilai lim 𝑥→ 𝜋 2 4(𝑥−𝜋)cos2x 𝜋(𝜋−2𝑥) tan(𝑥− 𝜋 2 ) = ⋯. A. −2 D. 1 B. −1 E. 2 C. 0 Pembahasan Misalkan 𝑦 = 𝑥 − 𝜋 2 maka 𝑥 = 𝜋 2 + 𝑦 lim 𝑥→ 𝜋 2 4(𝑥 − 𝜋)cos2 x 𝜋(𝜋 − 2𝑥) tan (𝑥 − 𝜋 2 ) = lim 𝑦→0 4 ( 𝜋 2 + 𝑦 − 𝜋) cos2 ( 𝜋 2 + 𝑦) 𝜋 (𝜋 − 2. ( 𝜋 2 + 𝑦)) tan 𝑦
- 43. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 43 SMAN 12 MAKASSAR = lim 𝑦→0 4 (𝑦 − 𝜋 2 ) cos2 ( 𝜋 2 + 𝑦) 𝜋(−𝑦) tan 𝑦 = lim 𝑦→0 (4𝑦 − 2𝜋)(− sin 𝑦)2 𝜋(−𝑦) tan 𝑦 = lim 𝑦→0 (4𝑦 − 2𝜋) −𝜋 . lim 𝑦→0 sin 𝑦 𝑦 . lim 𝑦→0 sin 𝑦 tan 𝑦 = (4.0 − 2𝜋) −𝜋 . 1. 1 1 = −2𝜋 −𝜋 = 2 Jawaban E 105. Nilai lim 𝑥→𝑦 tan 𝑥−tan 𝑦 (1− 𝑥 𝑦 )(1+tan 𝑥 tan 𝑦) = ⋯. A. y D. −1 B. 1 E. – 𝑦 C. 0 Pembahasan lim 𝑥→𝑦 tan 𝑥 − tan 𝑦 (1 − 𝑥 𝑦 ) (1 + tan 𝑥 tan 𝑦) = lim 𝑥→𝑦 tan(𝑥 − 𝑦) (1 − 𝑥 𝑦 ) = lim 𝑥→𝑦 tan(𝑥 − 𝑦) ( 𝑦 − 𝑥 𝑦 ) = lim 𝑥→𝑦 tan(𝑥 − 𝑦) 1 𝑦 (𝑦 − 𝑥) = 𝑦lim 𝑥→𝑦 tan(𝑥 − 𝑦) (𝑦 − 𝑥) = 𝑦lim 𝑥→𝑦 tan(𝑥 − 𝑦) −(𝑥 − 𝑦) = −𝑦lim 𝑥→𝑦 tan(𝑥 − 𝑦) (𝑥 − 𝑦) = −𝑦. 1 = −𝑦 Jawaban: E 106. Nilai lim 𝑎→𝑏 tan 𝑎−tan 𝑏 1+(1− 𝑎 𝑏 ) tan 𝑎 tan 𝑏− 𝑎 𝑏 = ⋯. A. b D. −1 B. 1 E. – 𝑏 C. 0 Pembahasan lim 𝑎→𝑏 tan 𝑎 − tan 𝑏 1 + (1 − 𝑎 𝑏 ) tan 𝑎 tan 𝑏 − 𝑎 𝑏 = lim 𝑎→𝑏 tan 𝑎 − tan 𝑏 (1 − 𝑎 𝑏 ) + (1 − 𝑎 𝑏 ) tan 𝑎 tan 𝑏
- 44. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 44 SMAN 12 MAKASSAR = lim 𝑎→𝑏 tan 𝑎 − tan 𝑏 (1 − 𝑎 𝑏 ) (1 + tan 𝑎 tan 𝑏) = lim 𝑎→𝑏 tan(𝑎 − 𝑏) (1 − 𝑎 𝑏 ) = lim 𝑎→𝑏 tan(𝑎 − 𝑏) ( 𝑏 − 𝑎 𝑏 ) = lim 𝑎→𝑏 tan(𝑎 − 𝑏) 1 𝑏 (𝑏 − 𝑎) = 𝑏lim 𝑎→𝑏 tan(𝑎 − 𝑏) (𝑏 − 𝑎) = 𝑏lim 𝑎→𝑏 tan(𝑎 − 𝑏) −(𝑎 − 𝑏) = −𝑏lim 𝑎→𝑏 tan(𝑎 − 𝑏) (𝑎 − 𝑏) = −𝑏. 1 = −𝑏 Jawaban E 107. Nilai lim 𝑥→ 𝜋 2 2𝑥− 𝜋 cos 𝑥 = ⋯. A. 4 D. −2 B. 2 E. −4 C. 0 Pembahasan Misalkan 𝑦 = 2𝑥 − 𝜋 sehingga 𝑥 = 𝜋 2 + 𝑦 2 Jika 𝑥 → 𝜋 2 maka 𝑦 → 0 lim 𝑥→ 𝜋 2 2𝑥 − 𝜋 cos 𝑥 = lim 𝑦→0 𝑦 cos ( 𝜋 2 + 𝑦 2 ) = lim 𝑦→0 𝑦 −sin 𝑦 2 = 1 − 1 2 = −2 Jawaban D 108. Nilai lim 𝑥→1 sin 𝜋𝑥 𝑥−1 = ⋯. A. −𝜋 D. 1 B. −1 E. 𝜋 C. 0 Pembahasan Misalkan 𝑦 = 𝑥 − 1 sehingga 𝑥 = 𝑦 + 1 Jika 𝑥 → 1 maka 𝑦 → 0 lim 𝑥→1 sin 𝜋𝑥 𝑥 − 1 = lim 𝑦→0 sin 𝜋(𝑦 + 1) 𝑦 = lim 𝑦→0 sin(𝜋𝑦 + 𝜋) 𝑦 = lim 𝑦→0 −sin 𝜋𝑦 𝑦 = −𝜋 Jawaban A
- 45. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 45 SMAN 12 MAKASSAR 109. Nilai lim 𝑥→−2 tan 𝜋𝑥 𝑥+2 = ⋯. A. −𝜋 D. 1 B. −1 E. 𝜋 C. 0 Pembahasan Sifat yang digunkan: tan(2𝜋 − 𝑎) = − tan 𝑎 Misalkan 𝑦 = 𝑥 + 2, sehingga 𝑥 = 𝑦 − 2 Jika 𝑥 → −2 maka 𝑦 → −2 + 2 = 0 lim 𝑥→−2 tan 𝜋𝑥 𝑥 + 2 = lim 𝑦→0 tan 𝜋(𝑦 − 2) 𝑦 = lim 𝑦→0 tan(𝜋𝑦 − 2𝜋) 𝑦 = lim 𝑦→0 tan 𝜋𝑦 𝑦 = 𝜋 Jawaban E 110. Nilai dari lim 𝑥→𝜋 1+cos 𝑥 (𝑥−𝜋)2 =…. A. −0,50 D. 0,25 B. −0,25 E. 0,50 C. 0 Pembahasan Misalkan 𝑦 = 𝑥 − 𝜋, sehingga 𝑥 = 𝜋 + 𝑦 Jika 𝑥 → 𝜋 maka 𝑦 → 𝜋 − 𝜋 = 0 lim 𝑥→𝜋 1 + cos 𝑥 (𝑥 − 𝜋)2 = lim 𝑦→0 1 + cos(𝜋 + 𝑦) 𝑦2 = lim 𝑦→0 1 − cos 𝑦 𝑦2 = lim 𝑦→0 2 𝑠𝑖𝑛2 1 2 𝑦 𝑦2 = 2 (lim 𝑦→0 sin 1 2 𝑦 𝑦 ) 2 = 2. 1 2 2 = 2 4 = 0,50 Jawaban E 111. Nilai lim 𝑥→ 𝜋 2 sin 2𝑥 𝑥− 𝜋 2 =... A. −2 D. 1 B. −1 E. 2 C. 0
- 46. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 46 SMAN 12 MAKASSAR Pembahasan Misalkan 𝑥 − 𝜋 2 = 𝑦 atau 𝑥 = 𝜋 2 + 𝑦 sin 2𝑥 = sin 2 ( 𝜋 2 + 𝑦) = sin(𝜋 + 𝑦) = − sin 𝑦 lim 𝑥→ 𝜋 2 sin 2𝑥 𝑥 − 𝜋 2 = lim 𝑦→0 − sin 𝑦 𝑦 = − lim 𝑦→0 sin 𝑦 𝑦 = −1 Jawaban B 112. Nilai lim 𝑥→𝜋 𝑥−𝜋 sin 𝑥 =.... A. −2 D. 1 B. −1 E. 2 C. 0 Pembahasan Misalkan 𝑥 − 𝜋 = 𝑦 atau 𝑥 = 𝜋 + 𝑦 sin 𝑥 = sin(𝜋 + 𝑦) = − sin 𝑦 Jika 𝑥 → 𝜋 maka 𝑦 → 0 lim 𝑥→𝜋 𝑥 − 𝜋 sin 𝑥 = lim 𝑦→0 𝑦 −sin 𝑦 = − lim 𝑦→0 𝑦 sin 𝑦 = −1 Jawaban B 113. Nilai lim 𝑥→ 𝜋 2 1−sin 𝑥 (𝜋−2𝑥)2 =.... A. 8 D. 1 2 B. 4 E. 1 8 C. 2 Pembahasan Misalkan 𝜋 − 2𝑥 = 𝑦 sehingga 𝑥 = 𝜋 2 − 𝑦 2 sin 𝑥 = sin ( 𝜋 2 − 𝑦 2 ) = cos 𝑦 2 Jika 𝑥 → 𝜋 2 maka 𝑦 → 0 lim 𝑥→ 𝜋 2 1 − sin 𝑥 (𝜋 − 2𝑥)2 = lim 𝑦→0 1 − cos 𝑦 2 𝑦2 = lim 𝑦→0 2 𝑠𝑖𝑛2 𝑦 4 𝑦2 = 2lim 𝑦→0 sin 𝑦 4 𝑦 . lim 𝑦→0 sin 𝑦 4 𝑦 = 2. 1 4 . 1 4 = 1 8 Jawaban E
- 47. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 47 SMAN 12 MAKASSAR 114. Nilai lim 𝑥→1 (1 − 𝑥) tan ( 𝜋𝑥 2 )=…. A. 𝜋 2 D. 𝜋 B. 2 𝜋 E. 0 C. 3 𝜋 Pembahasan Misalkan (1 − 𝑥) = 𝑦 lim 𝑥→1 (1 − 𝑥) tan ( 𝜋𝑥 2 ) = lim 𝑦→0 𝑦 tan ( 𝜋(1 − 𝑦) 2 ) = lim 𝑦→0 𝑦 tan ( 𝜋 2 − 𝜋 2 𝑦) = lim 𝑦→0 𝑦 cot ( 𝜋 2 𝑦) = lim 𝑦→0 𝑦 tan ( 𝜋 2 𝑦) = 1 𝜋 2 = 2 𝜋 Jawaban B 115. Nilai lim 𝑥→𝜋 1+cos 𝑥 𝑡𝑎𝑛2 𝑥 =…. A. −1 D. 1 2 B. − 1 2 E. 1 C. 0 Pembahasan Misalkan 𝑥 − 𝜋 = 𝑡 → 𝑥 = 𝜋 + 𝑡 Jika 𝑥 → 𝜋 maka 𝑡 → 0 lim 𝑥→𝜋 1 + cos 𝑥 𝑡𝑎𝑛2 𝑥 = lim 𝑡→0 1 + cos(𝜋 + 𝑡) 𝑡𝑎𝑛2(𝜋 + 𝑡) = lim 𝑡→0 1 − cos 𝑡 𝑡𝑎𝑛2 𝑡 = lim 𝑡→0 2 𝑠𝑖𝑛2 1 2 𝑡 𝑡𝑎𝑛2 𝑡 = 2 lim 𝑡→0 sin 1 2 𝑡 tan 𝑡 . lim 𝑡→0 sin 1 2 𝑡 tan 𝑡 = 2. 1 2 . 1 2 = 1 2 Jawaban D
- 48. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 48 SMAN 12 MAKASSAR 116. lim 𝑥→1 tan(𝑥2−1) (𝑥−1) =…. A. 2 D. −2 B. 1 2 E. − 1 2 C. 0 Pembahasan lim 𝑥→1 tan(𝑥2 − 1) (𝑥 − 1) = lim 𝑥→1 tan(𝑥2 − 1) (𝑥 − 1) = lim 𝑥→1 tan(𝑥2 − 1) (𝑥 − 1) . (𝑥 + 1) (𝑥 + 1) = lim 𝑥→1 tan(𝑥2 − 1) (𝑥2 − 1) . lim 𝑥→1 (𝑥 + 1) = 1. (1 + 1) = 2 Jawaban A 117. Nilai lim 𝑥→0 √1−cos 𝑥 𝑥 adalah …. A. −√2 D. 1 2 √2 B. − 1 2 √2 E. √2 C. 0 Pembahasan lim 𝑥→0 √1 − cos 𝑥 𝑥 = lim 𝑥→0 √1 − cos 𝑥 √𝑥2 = lim 𝑥→0 √ 1 − cos 𝑥 𝑥2 = √lim 𝑥→0 1 − cos 𝑥 𝑥2 = √ lim 𝑥→0 2 𝑠𝑖𝑛2 1 2 𝑥 𝑥2 = √ 2lim 𝑥→0 sin 1 2 𝑥 𝑥 . lim 𝑥→0 sin 1 2 𝑥 𝑥 = √2. 1 2 . 1 2 = 1 2 √2 Jawaban D
- 49. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 49 SMAN 12 MAKASSAR 118. Jika diketahui lim 𝑥→0 𝑎𝑥 sin 𝑥+𝑏 cos 𝑥−1 = 1, maka nilai 𝑎 dan 𝑏 yang memenuhi adalah …. A. 𝑎 = − 1 2 , 𝑏 = 0 D. 𝑎 = 1, 𝑏 = −1 B. 𝑎 = 1, 𝑏 = 1 E. . 𝑎 = 1, 𝑏 = 0 C. 𝑎 = 1 2 , 𝑏 = 0 Pembahasan Karena cos 𝑥 − 1 bernilai 0 untuk 𝑥 = 0 dan nilai limit 1, maka bagian pembilang harus bernilai 0 𝑎. 0. sin 0 + 𝑏 = 0 sehingga 𝑏 = 0 lim 𝑥→0 𝑎𝑥 sin 𝑥 + 𝑏 cos 𝑥 − 1 = 1 ⇔ lim 𝑥→0 𝑎𝑥 sin 𝑥 cos 𝑥 − 1 = 1 ⇔ lim 𝑥→0 𝑎𝑥 sin 𝑥 −2 𝑠𝑖𝑛2 1 2 𝑥 = 1 ⇔ lim 𝑥→0 𝑎𝑥 sin 1 2 𝑥 . lim 𝑥→0 sin 𝑥 sin 1 2 𝑥 = −2 ⇔ 𝑎 1 2 . 1 1 2 = −2 ⇔ 𝑎 = −2 × 1 4 = − 1 2 Jawaban A 119. Misalkan 𝛼 dan 𝛽 adalah akar-akar persamaan 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, maka lim 𝑥→𝛼 1−cos(𝑎𝑥2+𝑏𝑥+𝑐) (𝑥−𝛼)2 sama dengan …. A. 0 D. 𝛼2 2 (𝛼 + 𝛽)2 B. 1 2 (𝛼 − 𝛽)2 E. 𝛽2 2 (𝛼 − 𝛽)2 C. 𝛼2 2 (𝛼 − 𝛽)2 Pembahasan lim 𝑥→𝛼 1 − cos(𝑎𝑥2 + 𝑏𝑥 + 𝑐) (𝑥 − 𝛼)2 = lim 𝑥→𝛼 2 sin2 1 2 (𝑎𝑥2 + 𝑏𝑥 + 𝑐) (𝑥 − 𝛼)2 = lim 𝑥→𝛼 2 sin2 1 2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼)2 = 2 (lim 𝑥→𝛼 sin 1 2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼) ) 2 = 2 (lim 𝑥→𝛼 sin 1 2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼) . (𝑥 − 𝛽) (𝑥 − 𝛽) ) 2 = 2 (lim 𝑥→𝛼 (𝑥 − 𝛽)sin 1 2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼)(𝑥 − 𝛽) ) 2
- 50. Muhammad Arif,S.Pd,M.Pd. 120 LimitFungsi Trigonometri 50 SMAN 12 MAKASSAR = 2 (lim 𝑥→𝛼 (𝑥 − 𝛽)lim 𝑥→𝛼 sin 1 2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼)(𝑥 − 𝛽) ) 2 = 2 ((𝛼 − 𝛽). 1 2 ) 2 = 2. 1 4 (𝛼 − 𝛽)2 = 1 2 (𝛼 − 𝛽)2 Jawaban B 120. Jika 𝑓(𝑥) = cos 2𝑥 maka lim ℎ→0 𝑓(𝑥+2ℎ)−2𝑓(𝑥)+𝑓(𝑥−2ℎ) (2ℎ)2 = ⋯. A. 2 cos 2𝑥 D. −4 cos 2𝑥 B. −2 sin 2𝑥 E. 2 cos 4𝑥 C. 4 sin 2𝑥 Pembahasan o 𝑓(𝑥) = cos 2𝑥 o 𝑓(𝑥 + 2ℎ) = cos 2(𝑥 + 2ℎ) = cos(2𝑥 + 4ℎ) o 𝑓(𝑥 − 2ℎ) = cos 2(𝑥 − 2ℎ) = cos(2𝑥 − 4ℎ) o 𝑓(𝑥 + 2ℎ) − 𝑓(𝑥 − 2ℎ) = cos(2𝑥 + 4ℎ) − cos(2𝑥 − 4ℎ) 𝑓(𝑥 + 2ℎ) − 𝑓(𝑥 − 2ℎ) = 2 cos 1 2 (2𝑥 + 4ℎ + 2𝑥 − 4ℎ) cos 1 2 (2𝑥 + 4ℎ − 2𝑥 + 4ℎ) 𝑓(𝑥 + 2ℎ) − 𝑓(𝑥 − 2ℎ) = 2 cos 2𝑥 cos 4ℎ Sehingga lim ℎ→0 𝑓(𝑥 + 2ℎ) − 2𝑓(𝑥) + 𝑓(𝑥 − 2ℎ) (2ℎ)2 = lim ℎ→0 2 cos 2𝑥 cos 4ℎ − 2 cos 2𝑥 (2ℎ)2 = lim ℎ→0 2 cos 2𝑥 (cos 4ℎ − 1) 4ℎ2 = lim ℎ→0 2 cos 2𝑥 (−2. 𝑠𝑖𝑛2 2ℎ) 4. ℎ2 = lim ℎ→0 − cos 2𝑥 . sin 2 2ℎ ℎ2 = lim ℎ→0 − cos 2𝑥 lim ℎ→0 sin 2 2ℎ ℎ2 = − cos 2𝑥 lim ℎ→0 sin 2ℎ ℎ . lim ℎ→0 sin 2ℎ ℎ = − cos 2𝑥 . 2.2 = −4 cos 2𝑥 Jawaban D Kritik dan saran: [email protected]