Soal dan pembahasan limit fungsi trigonometri untuk materi matematika peminatan atau kalkulus.

1. Muhammad Arif,S.Pd,M.Pd.
120 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 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

23. Muhammad Arif,S.Pd,M.Pd.
120 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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 Limit Fungsi 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
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