120 SOAL DAN PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI

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SMAN 12 MAKASSAR

SOAL DAN PEMBAHASAN LIMIT FUNGSI TRIGONOMETRI

1. Nilai lim 𝑥→𝜋₃

2 tan 𝑥−sin 𝑥 cos 𝑥 = ⋯.

A. 3√3 D.3₄√3

B. 5₂√3 E.1₄√3

C. 3 2√3 Pembahasan

lim 𝑥→𝜋3

2 tan 𝑥 − sin 𝑥

cos 𝑥 =2 tan 𝜋3 − sin 𝜋 3 cos 𝜋3 = 2. √3 − 12√3₁

2

=

4√3 − √3 2 1 2 =4√3 − √3₁ = 3√3

Jawaban A 2. Nilai lim

𝑥→ 𝜋₄

sin 2𝑥

sin 𝑥+ cos 𝑥= ⋯.

A. √2 D. 0

B. 1₂√2 E. −1

C. 1

Pembahasan

lim 𝑥→ 𝜋4

sin 2𝑥

sin 𝑥 + cos 𝑥 = sin 2. 𝜋4 sin 𝜋4 + cos 𝜋4 = sin 𝜋2

sin 𝜋4 + cos 𝜋4

=₁ 1

2 √2 +12 √2 = 1

√2 =1_{2 √2}

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SMAN 12 MAKASSAR

3. Nilai lim 𝑥→ 𝜋₄

1−2 𝑠𝑖𝑛2𝑥 cos 𝑥−sin 𝑥= ⋯.

A. 0 D. √2

B. 1

2√2 E. ∞

C. 1

Pembahasan

lim 𝑥→ 𝜋4

1 − 2 𝑠𝑖𝑛2_𝑥

cos 𝑥 − sin 𝑥 = lim_{𝑥→ 𝜋4}

𝑐𝑜𝑠2_{𝑥 + 𝑠𝑖𝑛}2_{𝑥 − 2 𝑠𝑖𝑛}2_𝑥 cos 𝑥 − sin 𝑥

= lim 𝑥→ 𝜋4

𝑐𝑜𝑠2_{𝑥 − 𝑠𝑖𝑛}2_𝑥 cos 𝑥 − sin 𝑥 = lim

𝑥→ 𝜋4

(cos 𝑥 − sin 𝑥)(cos 𝑥 + sin 𝑥) cos 𝑥 − sin 𝑥

= lim

𝑥→ 𝜋4(cos 𝑥 + sin 𝑥) = cos𝜋_{4 + sin}𝜋₄

=1_{2 √2 +}1_{2 √2} = √2

Jawaban D

cos 2𝑥

cos 𝑥−sin 𝑥= ⋯.

A. −√2 D. 1

2√2

B. −1₂√2 E. √2

C. 0

Pembahasan

lim 𝑥→ 𝜋4

cos 2𝑥

cos 𝑥 − sin 𝑥 = lim_{𝑥→ 𝜋4}𝑐𝑜𝑠

2_{𝑥 − 𝑠𝑖𝑛}2_𝑥 cos 𝑥 − sin 𝑥 = lim

𝑥→ 𝜋4

(cos 𝑥 − sin 𝑥)(cos 𝑥 + sin 𝑥) cos 𝑥 − sin 𝑥

= lim

𝑥→ 𝜋4(cos 𝑥 + sin 𝑥) = cos𝜋_{4 + sin}𝜋₄

=1_{2 √2 +}1_{2 √2} = √2

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SMAN 12 MAKASSAR

1−2 sin 𝑥.cos 𝑥 sin 𝑥−cos 𝑥 = ⋯.

A. 1 D. 0

B. 1₂√2 E. −1

C. 1₂

Pembahasan

lim 𝑥→ 𝜋4

1 − 2 sin 𝑥 . cos 𝑥

sin 𝑥 − cos 𝑥 = lim𝑥→ 𝜋₄

sin2_x+cos2_{x−2 sin 𝑥.cos 𝑥}

sin 𝑥−cos 𝑥 ;karena sin2x + cos2x = 1

= lim 𝑥→ 𝜋4

(sin 𝑥 − cos 𝑥)2 sin 𝑥 − cos 𝑥 = lim

𝑥→ 𝜋4(sin 𝑥 − cos 𝑥) = sin𝜋_{4 − cos}𝜋₄

= 1_{2 √2 −}1_{2 √2} =0

Jawaban D 6. Nilai dari lim

𝑥→𝜋₈

𝑠𝑖𝑛2_{2𝑥−𝑐𝑜𝑠}2_2𝑥 sin 2𝑥−cos 2𝑥 = ….

A. 0 D. 1₂√2

B. 1₂ E. 1

C. √2

Pembahasan

lim 𝑥→𝜋8

𝑠𝑖𝑛2_{2𝑥 − 𝑐𝑜𝑠}2_2𝑥

sin 2𝑥 − cos 2𝑥 = lim_𝑥→𝜋8

(sin 2𝑥 − cos 2𝑥)(𝑠𝑖𝑛 2𝑥 + cos 2𝑥) sin 2𝑥 − cos 2𝑥

= lim

𝑥→𝜋8(𝑠𝑖𝑛 2𝑥 + cos 2𝑥) = (𝑠𝑖𝑛 2.𝜋_{8 + cos 2.}𝜋₈₎ = sin𝜋_{4 + cos}𝜋₄

= 1_{2 √2 +}1_{2 √2} = √2

Jawaban C

7. Nilai dari lim 𝑥→𝜋₂

𝑥 𝑐𝑜𝑡2_𝑥 1−sin 𝑥= ….

A. −2𝜋 D. 𝜋

B. – 𝜋 E. 2𝜋

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SMAN 12 MAKASSAR

9. Nilai lim 𝑥→ 0

sin 2𝑥 −2 sin 𝑥 𝑥3 = ⋯.

A. 3₂ D. −1

B. 1₂ 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 (−2𝑠𝑖𝑛 21

2 𝑥) 𝑥3

= −4 lim_{𝑥→ 0}sin 𝑥 . sin 12𝑥 .sin 1 2 𝑥 𝑥3

= −4 lim_{𝑥→ 0} sin 𝑥_{𝑥 lim}_{𝑥→ 0}sin 12𝑥_{𝑥 . lim}_{𝑥→ 0}sin 12𝑥_𝑥

= −4.1_{2 .}1₂ =−1

Jawaban D

10.Nilai lim 𝑥→𝜋₄

2−𝑐𝑠𝑐2_𝑥

1−cot 𝑥 adalah ….

A. – 2 D. 1

B. – 1 E. 2

C. 0

Pembahasan

lim 𝑥→𝜋4

2 − 𝑐𝑠𝑐2_𝑥

1 − cot 𝑥 = lim_𝑥→𝜋4

2 − (1 + 𝑐𝑜𝑡2_𝑥) 1 − cot 𝑥

= lim 𝑥→𝜋4

1 − 𝑐𝑜𝑡2_𝑥 1 − cot 𝑥 = lim

𝑥→𝜋4

(1 − cot 𝑥)(1 + cot 𝑥) 1 − cot 𝑥 = lim

𝑥→𝜋4(1 + cot 𝑥) = (1 + cot𝜋₄₎ = 1 + 1 =2

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SMAN 12 MAKASSAR

18. Nilai lim 𝑥→0

2𝑥 tan 𝑥 𝑡𝑎𝑛2𝑥

6 = ⋯.

A. 1₃ D. 36

B. 3 E. 72

C. 12

Pembahasan

lim 𝑥→0

2𝑥 tan 𝑥 𝑡𝑎𝑛2𝑥

6

= lim_𝑥→0 2𝑥 tan 𝑥6. lim𝑥→0

𝑡𝑎𝑛 𝑥 tan 𝑥6 =

2 1 6

.1₁ 6

= 12.6 = 72

Jawaban E

tan 2𝑥.tan 3𝑥 3𝑥2 = ⋯.

A. 0 D. 2

B. 2₃ E. 6

C. 3₂

Pembahasan

lim 𝑥→0

tan 2𝑥 . tan 3𝑥 3𝑥2 = lim_𝑥→0

tan 2𝑥 3𝑥 . lim𝑥→0

tan 3𝑥

𝑥 =

2

3 . 3 = 2 Jawaban D

sin 4𝑥−sin 2𝑥 6𝑥 = ⋯.

A. 1₆ D. 2₃

B. 1₃ E.1

C. 1₂

Pembahasan

lim 𝑥→0

sin 4𝑥 − sin 2𝑥

6𝑥 = lim𝑥→02 cos 12(4𝑥 + 2𝑥).sin 1

2 (4𝑥 − 2𝑥) 6𝑥

= lim_𝑥→02 cos 3𝑥 . sin 𝑥 6𝑥

=2_{6 lim}_𝑥→0cos 3𝑥 . lim_𝑥→0sin 𝑥_𝑥 =2₆. cos 0.1

=1_{3 . 1.1}

=1₃

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SMAN 12 MAKASSAR Atau dengan cara berikut

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SMAN 12 MAKASSAR

= 2

(√3.0 + 1). 1 (√1 + 1) =2

1 . 1 2 = 1 Jawaban E

38. Jika lim 𝑥→0

𝑥𝑎𝑠𝑖𝑛4𝑥

𝑠𝑖𝑛6_𝑥 = 1, nilai 𝑎yang memenuhi adalah ….

A. 1 D. 4

B. 2 E. 5

C. 3

Pembahasan

lim 𝑥→0

𝑥𝑎_𝑠𝑖𝑛4_𝑥 𝑠𝑖𝑛6_{𝑥 = lim}_𝑥→0

𝑥𝑎 𝑠𝑖𝑛2_{𝑥 . lim}_𝑥→0

𝑠𝑖𝑛4_𝑥

𝑠𝑖𝑛4_{𝑥 = lim}_𝑥→0 𝑥𝑎

𝑠𝑖𝑛2_{𝑥 . 1 = lim}_𝑥→0 𝑥𝑎 𝑠𝑖𝑛2_𝑥 lim

𝑥→0 𝑥𝑎

𝑠𝑖𝑛2_𝑥= 1 hanya terjadi jika nilai 𝑎 sama dengan pangkat dari sin 𝑥, yaitu 𝑎 = 2 Jawaban B

1−𝑐𝑜𝑠3𝑥 𝑥.tan 𝑥 =….

A. 1₂ D. 2

B. 1 E. 3

C. 3₂

Pembahasan

lim 𝑥→0

1 − 𝑐𝑜𝑠3_𝑥

𝑥. tan 𝑥 = lim𝑥→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→02 . 𝑠𝑖𝑛 21

2 𝑥

𝑥. tan 𝑥 lim𝑥→0(1 + cos 𝑥 + 𝑐𝑜𝑠 2_𝑥)

= 2lim_𝑥→0sin 12𝑥_{𝑥 lim}_𝑥→0sin 12𝑥_{tan 𝑥 . lim}_𝑥→0lim_𝑥→0(1 + cos 𝑥 + 𝑐𝑜𝑠2_𝑥)

= 2.1_{2 .}1_{2 .}(1 + cos 0 + 𝑐𝑜𝑠2₀₎

=1₂(1 + cos 0 + 𝑐𝑜𝑠2₀₎

=1₂(1 + 1 + 12₎

=3₂

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SMAN 12 MAKASSAR

Pembahasan

lim 𝑥→0

1 − 𝑐𝑜𝑠2_𝑥

𝑥2_{cot (𝑥 − 𝜋3)} = lim𝑥→0

𝑠𝑖𝑛2_𝑥 𝑥2_{cot (𝑥 − 𝜋3)}

= lim_𝑥→0sin 𝑥_{𝑥 . lim}_𝑥→0sin 𝑥_{𝑥 . lim}_𝑥→0 1 cot (𝑥 − 𝜋3) = 1.1. 1

cot (0 − 𝜋3)

= 1

cot (− 𝜋3) = − tan𝜋₃ = −√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 = lim𝑥→0

−𝑠𝑖𝑛2_𝑥 2 sin 2𝑥 tan 𝑥

= −1_{2 lim}_𝑥→0_{sin 2𝑥 . lim}sin 𝑥 _𝑥→0_{tan 𝑥}sin 𝑥

= −1_{2 .}1_{2 . 1}

Jawaban E

1−𝑐𝑜𝑠2(𝑥−1) 4𝑥2_−8𝑥+4 = ⋯.

A. 0 D.1

B. 1₄ E.2

C. 1₂

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→1sin(𝑥 − 1)_{(𝑥 − 1) . lim}_𝑥→1sin(𝑥 − 1)_{(𝑥 − 1)}

=1_{4 . 1.1}

=1₄

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SMAN 12 MAKASSAR

= 1_{2 .}3₁ 2

.3₁ 2 = 9₁

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 𝑠𝑖𝑛21 2 𝜃 𝜃2

= 2 lim_𝜃→0 𝑠𝑖𝑛 21

2 𝜃 𝜃2

= 2 lim_𝜃→0𝑠𝑖𝑛 12𝜃_{𝜃 . lim}_𝜃→0𝑠𝑖𝑛 12𝜃_𝜃

= 2.1_{2 .}1₂

= 1₂

Jawaban E

1−cos 8𝑥 4𝑥2 =….

A. 0 D. 4

B. 1 E. 8

C. 2

Pembahasan

lim 𝑥→0

1 − cos 8𝑥

4𝑥2 = lim_𝑥→02 𝑠𝑖𝑛 2_4𝑥 4𝑥2 = 2_{4 lim}_𝑥→0 𝑠𝑖𝑛_𝑥2₂4𝑥

= 1_{2 lim}_𝑥→0sin 4𝑥_{𝑥 . lim}_𝑥→0sin 4𝑥_𝑥

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SMAN 12 MAKASSAR

1−𝑐𝑜𝑠 2𝑥 1−cos 4𝑥 =…. A. −1₂

B. −1₄ C. 0 D. ₁₆1 E. 1₄

Pembahasan

lim 𝑥→0

1 − 𝑐𝑜𝑠 2𝑥

1 − cos 4𝑥 = limx→0

2 sin2_x 2 sin2_2x = lim_x→0_sin sin₂2_2xx

= lim_𝑥→0_{sin 2𝑥 . lim}sin 𝑥 _𝑥→0_{sin 2𝑥}sin 𝑥

= 1_{2 .}1₂

= 1₄ Jawaban E

1−𝑐𝑜𝑠 2𝑥 𝑥 𝑡𝑎𝑛𝑥 = ⋯.

A. −8 D.2

B. 0 E.4

C. 1

Pembahasan

lim 𝑥→0

1 − 𝑐𝑜𝑠 2𝑥

𝑥 𝑡𝑎𝑛𝑥 = lim𝑥→0

2𝑠𝑖𝑛2_𝑥 𝑥 𝑡𝑎𝑛𝑥 = 2 lim_𝑥→0sin 𝑥 sin 𝑥_{𝑥 𝑡𝑎𝑛𝑥}

= 2 lim_𝑥→0sin 𝑥_{𝑥 . lim}_𝑥→0_{tan 𝑥}sin 𝑥 = 2.1.1

= 2

Jawaban D

(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→02. 𝑠𝑖𝑛

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SMAN 12 MAKASSAR

= 2 lim_𝑥→0𝑠𝑖𝑛_𝑥₂22𝑥. sin 𝑥_{tan 3𝑥}

= 2 (lim_𝑥→0sin 2𝑥_{𝑥 )}2. lim_𝑥→0_{tan 3𝑥}sin 𝑥

= 2. 22_.1 3 =8₃

Jawaban D

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 = lim𝑥→0

−2𝑠𝑖𝑛2_2𝑥 𝑥 tan 2𝑥 = −2 lim_𝑥→0_{𝑥 tan 2𝑥}𝑠𝑖𝑛22𝑥

= −2. lim_𝑥→0sin 2𝑥 𝑥 . lim𝑥→0

sin 2𝑥 tan 2𝑥 = −2.2.2₂

= −4 Jawaban E

cos 6𝑥 −1 𝑥 sin1₂𝑥

A. 36 D.−9

B. 9 E. −36

C. 0

Pembahasan

lim 𝑥→0

cos 6𝑥 − 1

𝑥 sin 12𝑥 = lim𝑥→0

−2𝑠𝑖𝑛2_3𝑥 𝑥 sin 12𝑥 = −2 lim_𝑥→0 𝑠𝑖𝑛23𝑥

𝑥 sin 12𝑥

= −2. lim_𝑥→0sin 3𝑥_{𝑥 . lim}_𝑥→0sin 3𝑥 sin 12𝑥 = −2.3.3₁

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SMAN 12 MAKASSAR

4x cos 6𝑥 −4𝑥 (2𝑥)2_{.sin 5𝑥} = A. −18

5 D. 2

B. −₁₈5 E. 18₅

C. ₁₈5

Pembahasan

lim 𝑥→0

4x cos 6𝑥 − 4𝑥

(2𝑥)2_{. sin 5𝑥} = lim_𝑥→04𝑥(cos 6𝑥 − 1)_(2𝑥)2_{. sin 5𝑥} = lim_𝑥→04𝑥(−2 𝑠𝑖𝑛_(2𝑥)₂_{. sin 5𝑥}23𝑥)

= lim_𝑥→0−8𝑥. sin 3𝑥 sin 3𝑥_(2𝑥)₂_{. sin 5𝑥}

= lim_𝑥→0_{sin 5𝑥 . lim}−8𝑥 _𝑥→0sin 3𝑥_{2𝑥 . lim}_𝑥→0sin 3𝑥_2𝑥

=−8_{5 .}3_{2 .}3₂

= −18₅ Jawaban A

58.Jika diketahui 𝑚 = lim 𝑥→0

cos 𝑥−1

cos 2𝑥−1 dan 𝑛 = lim𝑥→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 𝑠𝑖𝑛 21

2 𝑥 −2 𝑠𝑖𝑛2_𝑥

𝑚 = lim_𝑥→0 𝑠𝑖𝑛 21

2 𝑥 𝑠𝑖𝑛2_𝑥

𝑚 = (lim_𝑥→0 𝑠𝑖𝑛 12𝑥_{𝑠𝑖𝑛𝑥 )} 2

𝑚 = (1₂₎2

𝑚 =1₄

𝑛 = lim_𝑥→2[ 1 𝑥 − 2 −

4 𝑥2 _{− 4]} 𝑛 = lim_𝑥→2[_𝑥𝑥 + 2₂ _{− 4 −}_𝑥₂4_{− 4]}

𝑛 = lim_𝑥→2[𝑥 + 2 − 4_𝑥₂_{− 4 ]}

𝑛 = lim_𝑥→2[_𝑥𝑥 − 2₂ _{− 4]}

𝑛 = lim_𝑥→2_{(𝑥 − 2)(𝑥 + 2)}𝑥 − 2

𝑛 = lim_𝑥→2 1 (𝑥 + 2) 𝑛 =_{2 + 2}1

𝑛 =1₄

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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_{sin(x + 2) . lim}(𝑥 + 2) _𝑥→−2tan(𝑥 + 2)_{sin(x + 2)} = (−2 − 2).1.1

= −4 Jawaban A

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) = lim𝑥→3 𝑥. lim𝑥→3

tan(2𝑥 − 6) sin(𝑥 − 3) = lim_𝑥→3 𝑥. lim_𝑥→3tan 2(𝑥 − 3)_{sin(𝑥 − 3)} = 3.2

=6 Jawaban A

cos 𝑥−cos 5𝑥 1−cos 4𝑥 = ⋯. A. 1

3 D.2

B. 1₂ E.3

C. 3₂

Pembahasan

lim 𝑥→0

cos 𝑥 − cos 5𝑥

1 − cos 4𝑥 = lim𝑥→0

−2 sin 3𝑥 . sin(−2𝑥) 2 𝑠𝑖𝑛2_2𝑥 = 2_{2 lim}_𝑥→0sin 3𝑥 . sin 2𝑥_𝑠𝑖𝑛₂_2𝑥

= lim_𝑥→0sin 3𝑥 . sin 2𝑥_𝑠𝑖𝑛₂_2𝑥

= lim_𝑥→0sin 3𝑥 . sin 2𝑥_{sin 2𝑥 . sin 2𝑥}

= lim_𝑥→0sin 3𝑥_{sin 2𝑥}

Rumus

cos 𝐴 − cos 𝐵 = −2𝑠𝑖𝑛1

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SMAN 12 MAKASSAR

= 3₂

Jawaban C

62. _𝑥→0lim_{cos 2𝑥−cos 7𝑥}𝑥.tan 5𝑥 = ⋯.

A. 2₉ D. −1₉

B. 1₉ E. −2₉

C. 0

Pembahasan

lim 𝑥→0

𝑥. tan 5𝑥

cos 2𝑥 − cos 7𝑥 = lim𝑥→0

𝑥. tan 5𝑥 −2 sin 92𝑥 sin(−52) 𝑥 =1_{2 lim}_𝑥→0 𝑥. tan 5𝑥

sin 92𝑥 sin52 𝑥 =1_{2 lim}_𝑥→0 𝑥

sin 92𝑥. lim𝑥→0 tan 5𝑥 sin 52𝑥 =1_{2 .}1₉

2 .5₅

2 =2₉ Jawaban A

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 = lim𝑥→0

2 𝑠𝑖𝑛2_4𝑥 sin 2𝑥 tan 2𝑥 = lim_𝑥→0sin 4𝑥 . sin 4𝑥_{sin 2𝑥 tan 2𝑥}

= lim_𝑥→0sin 4𝑥_{sin 2𝑥 . lim}_𝑥→0_{tan 2𝑥}sin 4𝑥

= 4_{2 .}4₂ = 4 Jawaban D

1−2𝑠𝑖𝑛2_{𝑥−𝑐𝑜𝑠}3_2𝑥 5𝑥2 = ⋯. A. 4

25 D.

4 5

B. 2₅ E.1

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SMAN 12 MAKASSAR

Pembahasan

lim 𝑥→0

1 − 2𝑠𝑖𝑛2_{𝑥 − 𝑐𝑜𝑠}3_2𝑥

5𝑥2 = lim_x→0

cos 2x − cos3_2x 5x2

= lim_x→0cos 2x (1 − cos_5x₂ 22x)

= lim_x→0cos 2x sin_5x₂ 22x = 1₅lim_x→0cos 2𝑥. (lim x→0

sin 2𝑥 𝑥 )

2

= 1_{5 . cos 2.0 .}(2)2

= 1_{5 . 1.4}

= 4₅

karena 𝟏 − 𝟐𝒔𝒊𝒏𝟐𝒙 = 𝐜𝐨𝐬 𝟐𝒙

Jawaban D 65. Nilai lim

𝑥→0

1−cos2x−cos x sin2x 𝑥4 = ⋯.

A. −1 D. 1₂

B. 0 E. 1

C. 1₄

Pembahasan

lim 𝑥→0

1 − cos2_{x − cos x sin}2_x

𝑥4 = lim_𝑥→0

sin2_{x − cos x sin}2_x 𝑥4

= lim_𝑥→0sin2x(1 − cos x) _𝑥₄

= lim_𝑥→0sin

2_{x. 2 𝑠𝑖𝑛}21 2 x 𝑥4

= 2 lim_𝑥→0sin

2_{x. 𝑠𝑖𝑛}21 2 x 𝑥4

= 2 lim_𝑥→0sin_𝑥2₂x . lim_𝑥→0sin 21

2 x 𝑥2

= 2 (lim_𝑥→0sin 𝑥 _{𝑥 )}2. (lim_𝑥→0sin 12𝑥_{𝑥 )} 2

= 2. 12_{. (}1 2)

2

=2.1₄

= 2₄

(26)

(27)

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SMAN 12 MAKASSAR

= 2lim_𝑥→0sin 𝑥_{𝑥 . (lim}_𝑥→0sin 12𝑥_{𝑥 )} 2

. lim_𝑥→0 1

cos 𝑥 (√1 + tan 𝑥 + √1 + sin 𝑥)

= 2.1. (1₂₎2. 1

cos 0(√1 + tan 0 +√1 + sin 0) = 2.1_{4 .} 1

1(√1 + √1) = 2₈

= 1 4 Jawaban D

√1+sin 𝑥−√1−sin 𝑥

𝑥 = ⋯.

A. −1 D. √2

B. −1

4 E.1

C. 1₄√2 Pembahasan

lim 𝑥→0

√1 + sin 𝑥 − √1 − sin 𝑥

𝑥 = lim𝑥→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→02 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

=2₂ = 1 Jawaban E

(1−√cos 𝑥) cot 𝑥 𝑥 = ⋯.

A. −1₂ D. 1₄

B. −1₄ E. 1₂

C. 0

Pembahasan

lim 𝑥→0

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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 21

2 x 𝑥(1 + √cos 𝑥) tan 𝑥

= lim_𝑥→02 sin 12𝑥_𝑥 . lim_𝑥→0sin 12𝑥_{tan 𝑥 . lim}_𝑥→0 1 (1 + √cos 𝑥) = 2.1_{2 .}1_{2 .} 1

1 + √cos 0 =1_{2 .}1₂

=1₄

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) = lim𝑥→3

(𝑥 − 3)(𝑥 + 3) sin(𝑥 − 3)

= lim_𝑥→3_{sin(𝑥 − 3) . lim}(𝑥 − 3) _𝑥→3(𝑥 + 3) = 1. (3 + 3)

= 6 Jawaban C

(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_{+ 3𝑥}2 _{+ 2𝑥} = lim_𝑥→0

(𝑥2_{− 1) sin 6𝑥} 𝑥(𝑥2_{+ 3𝑥 + 2)}

= lim_𝑥→1 _(𝑥₂(𝑥_{+ 3𝑥 + 2) . lim}2− 1) _𝑥→1 sin 6𝑥_𝑥

=₍₀₂(0_{+ 3.0 + 2) . 6}2− 1)

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SMAN 12 MAKASSAR

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)𝑥}2_{+ 𝑎𝑥}

(𝑥2_{− 𝑎) tan(𝑥 − 1)} = lim_𝑥→1

𝑥(𝑥2_{− (𝑎 + 1)𝑥 + 𝑎)} (𝑥2_{− 𝑎) tan(𝑥 − 1)} = lim_𝑥→1_(𝑥𝑥(𝑥 − 𝑎)(𝑥 − 1)₂_{− 𝑎) tan(𝑥 − 1)}

= lim_𝑥→1𝑥(𝑥 − 𝑎)_(𝑥₂_{− 𝑎) . lim}_𝑥→1_{tan(𝑥 − 1)}(𝑥 − 1)

=1(1 − 𝑎)₍₁₂_{− 𝑎) . 1}

=1 − 𝑎_{1 − 𝑎} = 1

Jawaban A

(𝑥2−1) sin 2(𝑥−1) −2 𝑠𝑖𝑛2_(𝑥−1) = ⋯.

A. −2 D.−1

4

B. −1 E.0

C. −1₂ 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_{sin(𝑥 − 1) . lim}(𝑥 − 1) _𝑥→1sin2(𝑥 − 1)_{sin(𝑥 − 1) . lim}_𝑥→1(𝑥 + 1)

=−1₂. 1.2. (1 + 1) = −2

Jawaban A

2 𝑠𝑖𝑛2𝑥 2 𝑥 sin 𝑥 = ⋯.

A. 0 D. 2

B. 1₄ E. 4

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Pembahasan

lim 𝑥→0

2 𝑠𝑖𝑛2𝑥 2

𝑥 sin 𝑥 = 2 lim𝑥→0sin 𝑥2.sin 𝑥 2 𝑥 sin 𝑥 = 2 lim_𝑥→0sin 𝑥2_{𝑥 . lim}_𝑥→0sin 𝑥2_{sin 𝑥}

= 2.1_{2 .}1₂

=1₂

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 = lim𝑥→0

sin 4𝑥 1 − √1 − 𝑥 .

1 + √1 − 𝑥 1 + √1 − 𝑥 = lim_𝑥→0sin 4𝑥(1 + √1 − 𝑥)_{1 − (1 − 𝑥)}

= lim_𝑥→0sin 4𝑥(1 + √1 − 𝑥)_𝑥

= lim_𝑥→0sin 4𝑥_{𝑥 . lim}_𝑥→0(1 + √1 − 𝑥) = 4. (1 + √1 − 0)

= 4(1 + 1) =8

Jawaban A 79. Nilai lim

𝑥→𝜋₃

tan(3𝑥−𝜋) cos 2𝑥 sin(3𝑥−𝜋) = ⋯.

A. −1₂ D.1₂√3

B. 1₂ E.3₂

C. 1₂√2 Pembahasan

lim 𝑥→𝜋3

tan(3𝑥 − 𝜋) cos 2𝑥

sin(3𝑥 − 𝜋) = lim_𝑥→𝜋3

tan(3𝑥 − 𝜋)

sin(3𝑥 − 𝜋) . lim_𝑥→𝜋3cos 2𝑥 = 1. cos (2.𝜋₃₎

= cos (2𝜋_{3 )}

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80. Nilai dari lim 𝑥→2

(𝑥−2) cos(𝜋𝑥−2𝜋) tan(2𝜋𝑥−4𝜋) =….

A. 2𝜋 D. 1_𝜋

B. 𝜋 E. _2𝜋1

C. 0

Pembahasan

lim 𝑥→2

(𝑥 − 2) cos(𝜋𝑥 − 2𝜋)

tan(2𝜋𝑥 − 4𝜋) = lim𝑥→2

(𝑥 − 2) cos 𝜋(𝑥 − 2) tan2𝜋(𝑥 − 2)

= lim_𝑥→2_{tan2𝜋(𝑥 − 2) . lim}(𝑥 − 2) _𝑥→2cos 𝜋(𝑥 − 2)

= _{2𝜋 . cos 𝜋}1 (2 − 2)

= 1

2𝜋 . cos 0 = _{2𝜋 . 1}1

= _2𝜋1 Jawaban E

81.Nilai lim 𝑥→−3

𝑥2_+6𝑥+9

2−2 cos(2𝑥+6)= ⋯.

A. 3 D.1₃

B. 1 E. 1₄

C. 1 2

Pembahasan

lim 𝑥→−3

𝑥2_{+ 6𝑥 + 9}

2 − 2 cos(2𝑥 + 6) = lim𝑥→−3

(𝑥 + 3)2 2(1 − cos(2𝑥 + 6)) =1_{2 lim}_𝑥→−3 _{(1 − cos 2(𝑥 + 3))}(𝑥 + 3)2

=1_{2 lim}_𝑥→−3 _{2 sin}(𝑥 + 3)₂_{(𝑥 + 3)}2

=1_{4 lim}_𝑥→−3 _sin(𝑥 + 3)₂_{(𝑥 + 3)}2

=1_{4 { lim}_𝑥→−3 _{sin(𝑥 + 3)}}(𝑥 + 3) 2

=1_{4 . 1}2

=1₄ Jawaban E

82.Nilai lim 𝑥→−2

2−2 cos(𝑥+2) 𝑥2_+4𝑥+4 =….

A. 4 D. 1

B. 2 E. 1₂

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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→−22. 𝑠𝑖𝑛 21

2 (𝑥 + 2) (𝑥 + 2)2

= 4 . { lim_𝑥→−2sin 12(𝑥 + 2)_{(𝑥 + 2) }} 2

= 4 . {1 2}

2

= 4.1₄ = 1 Jawaban D

tan(𝑥−1) sin(1−√𝑥) 𝑥2_−2𝑥+1 = ⋯.

A. −1 D. 1₂

B. −1₂ E. 1

C. 0

Pembahasan

lim 𝑥→1

tan(𝑥 − 1) sin(1 − √𝑥)

𝑥2_{− 2𝑥 + 1} = lim_𝑥→1tan(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→1tan(𝑥 − 1)_{(𝑥 − 1) . lim}_𝑥→1sin(1 − √𝑥)

(√𝑥 − 1) . lim𝑥→1 1 (√𝑥 + 1) = 1. (−1). 1

(√1 + 1) =−1₂

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

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SMAN 12 MAKASSAR

= (1 + 2). 1 = 3

Jawaban B

(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_{(2𝑥 + 1) . lim}(𝑥 + 2) _𝑥→4tan(𝑥 − 4)_{(𝑥 − 4)}

=_{(2.4 + 1) .1}(4 + 2)

=6₉

=2₃ 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→2tan(𝑥 − 2)_{(𝑥 − 2)}

=(2.2 + 1)_{(2 + 2)}

=5₄ =1,25 Jawaban A

(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

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SMAN 12 MAKASSAR

= lim_𝑥→1(3𝑥 + 1)_{(𝑥 + 3) . lim}_𝑥→1sin(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 + 1)}(𝑡 − 3)₂. lim_𝑡→2sin(𝑡 − 2)_{(𝑡 − 2)}

=_{(2 + 1)}(2 − 3)₂. 1

= −1₉ Jawaban D

1−1_𝑥

sin 𝜋(𝑥−1)=…. A. 2

𝜋 D. −

1 𝜋 B. 1

𝜋 E. −

2 𝜋 C. 0

Pembahasan

lim

𝑥→1sin 𝜋(𝑥 − 1)1 − 1𝑥 = lim𝑥→1

𝑥 − 1 𝑥 sin 𝜋(𝑥 − 1) = lim_𝑥→1_{x. sin 𝜋(𝑥 − 1)}(𝑥 − 1)

= lim_𝑥→11_{𝑥 . lim}_𝑥→1_{sin 𝜋(𝑥 − 1)}(𝑥 − 1)

=1_{1 .}1_𝜋

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90. Nilai lim 𝑥→ 𝜋

sin(𝑥−𝜋)

2(𝑥−𝜋)+tan(𝑥−𝜋)= ⋯.

A. −1₂ D. 1₃

B. −1₄ E. 2₅

C. 1₄

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) = lim𝑦→ 0

sin 𝑦 2𝑦 + tan 𝑦

= lim_{𝑦→ 0}

sin 𝑦 𝑦 2𝑦

𝑦 +tan 𝑦𝑦

= 𝑦→ 0lim sin 𝑦

𝑦 lim

𝑦→ 02 + lim𝑦→ 0 tan 𝑦

𝑦 =_{2 + 1}1

=1₃ Jawaban D

91. Nilai lim 𝑥→𝜋₃

sin(𝑥−𝜋₃)+sin 5(𝑥−𝜋₃) 6(𝑥−𝜋₃) = ⋯.

A. 1 D.3

B. 2 E.7₂

C. 5 2

Pembahasan Misalkan 𝑥 −𝜋₃ = 𝑦

Jika 𝑥 →𝜋₃ maka 𝑦 →𝜋₃−𝜋₃ = 0

lim 𝑥→𝜋3

sin (𝑥 − 𝜋3) + sin5(𝑥 −𝜋3) 6 (𝑥 − 𝜋3)

= lim_𝑦→0sin 𝑦 + sin 5𝑦_6𝑦

= lim_𝑦→0sin 𝑦_{6𝑦 + lim}_𝑦→0sin 5𝑦_6𝑦

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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 = lim𝑥→𝑎

𝑥 − 𝑎

sin(𝑥 − 𝑎) − 2(𝑥 − 𝑎) = lim_𝑦→0_{sin 𝑦 − 2𝑦}𝑦

= lim_𝑦→0 𝑦 𝑦 sin 𝑦

𝑦 −2𝑦𝑦

= 𝑦→0lim1 lim

𝑦→0 sin 𝑦

𝑦 − lim𝑦→02 =_{1 − 2}1

= −1 Jawaban D

93.Jika diketahui lim 𝑥→0

tan 𝑥

𝑥 = 1, maka 𝑥→𝑎lim

𝑥−𝑎

tan(𝑥−𝑎)+3𝑥−3𝑎=….

A. 0 D. 1₂

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 = lim𝑥→𝑎

𝑥 − 𝑎

tan(𝑥 − 𝑎) + 3(𝑥 − 𝑎) = lim_𝑦→0_{tan 𝑦 + 3𝑦}𝑦

= lim_𝑦→0 𝑦 𝑦 tan 𝑦

𝑦 +3𝑦𝑦

= 𝑦→0lim1 lim

𝑦→0 tan 𝑦

𝑦 + lim𝑦→03 = _{1 + 3}1

(38)

SMAN 12 MAKASSAR

1−cos 𝑥

cos 3𝑥−cos 𝑥= ⋯.

A. −1₈ D. 1₄

B. −1₄ E. 1₈

lim 𝑥→0

1 − cos 𝑥

cos 3𝑥 − cos 𝑥 = lim_𝑥→0 2 𝑠𝑖𝑛 21

2 𝑥 −2 sin 2𝑥 . 𝑠𝑖𝑛𝑥

= lim_𝑥→0sin 12𝑥 .sin 1 2 𝑥 − sin 2𝑥 . 𝑠𝑖𝑛𝑥

= lim_𝑥→0_{− sin 2𝑥 . lim}sin 12𝑥 _𝑥→0sin 12𝑥_{sin 𝑥}

= − 1 2 2 .

1 2 = −1₈ Jawaban A

cos 𝑥−cos 5𝑥 12𝑥 tan 2𝑥 = ⋯.

A. 1₆ D.−1₆

B. 1₂ E. −₁₂1

lim 𝑥→0

12𝑥 tan 2𝑥 = lim𝑥→0

−2 sin 3𝑥 sin(−2𝑥) 12𝑥 tan 2𝑥 =_{12 lim}2 _𝑥→0sin 3𝑥 sin 2𝑥_{𝑥 tan 2𝑥}

=1_{6 lim}_𝑥→0sin 3𝑥_{𝑥 . lim}_𝑥→0_{tan 2𝑥}sin 2𝑥

=1_{6 . 3.}2₂

=1₂ Jawaban B

96.Jika diketahui lim 𝑥→0

sin 𝑥

𝑥 = 1, maka 𝑥→0 lim

cos 𝑥−cos 2𝑥 𝑥2 =….

A. 1₂ D. 3₂

B. 2

3 E. 2

(39)

SMAN 12 MAKASSAR

Pembahasan

lim 𝑥→0

𝑥2 _{= lim}

𝑥→0 2 sin 32𝑥 sin 1 2 𝑥 𝑥2

= 2. lim_𝑥→0sin 32𝑥_{𝑥 . lim}_𝑥→0sin 12𝑥_𝑥

= 2.3_{2 .}1₂

=3₂ Jawaban D

97. _𝑎→0limcos 𝑚𝛼−cos 𝑛𝛼_𝛼₂ =….

A. 𝑚−𝑛₂ D. 𝑚+𝑛₂

B. 𝑚2₂−𝑛2 E. 𝑛2−𝑚₂ 2

C. 𝑚2₂+𝑛2 Pembahasan

lim 𝑎→0

cos 𝑚𝛼 − cos 𝑛𝛼 𝛼2

= lim_𝑎→0−2 sin (𝑚𝛼 + 𝑛𝛼)2 _𝛼₂. sin (𝑚𝛼 − 𝑛𝛼)2

= lim_𝑎→0−2 sin (𝑚𝛼 + 𝑛𝛼)_𝛼 2 . lim_𝑎→0sin (𝑚𝛼 − 𝑛𝛼)_𝛼2

= lim_𝑎→0−2 sin 𝛼(𝑚 + 𝑛)_𝛼 2 . lim_𝑎→0sin 𝛼(𝑚 − 𝑛)_𝛼2 = −2.(𝑚+𝑛)₂ .(𝑚−𝑛)₂

=(𝑚 + 𝑛)(𝑚 − 𝑛)₂

=𝑚2₂− 𝑛2 Jawaban B

98. Nilai dari  

 x x

x x

x cos

sin 5 sin 0 lim

…

A. 0 B. 1 C. 3 D. 4 E. 5

Pembahasan

lim 𝑥→0

sin 5𝑥 − sin 𝑥

𝑥 cos 𝑥 = lim𝑥→0

(40)

SMAN 12 MAKASSAR

= lim_𝑥→02 cos 3𝑥_{cos 𝑥 . lim}_𝑥→0sin 2𝑥_𝑥

= 2 cos 3.0_{cos 0 . 2}

= 2.1_{1 . 2} = 4 Jawaban D

sin 3𝑥−sin 3𝑥 cos 2𝑥 2𝑥3 = ⋯.

A. 4 D.1

B. 3 E.1₃

C. 2

Pembahasan

lim 𝑥→0

sin 3𝑥 − sin 3𝑥 cos 2𝑥

2𝑥3 = lim_𝑥→0

sin 3𝑥(1 − cos 2𝑥) 2𝑥3

= lim_𝑥→0sin 3𝑥. 2 𝑠𝑖𝑛_2𝑥₃ 2𝑥

= lim_𝑥→0sin 3𝑥. 𝑠𝑖𝑛_𝑥₃ 2𝑥

= lim_𝑥→0sin 3𝑥_{𝑥 . lim}_𝑥→0sin 𝑥_{𝑥 . lim}_𝑥→0sin 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→0sin 𝜃 ( 1cos 𝜃 − 1)_𝜃₃

= lim_𝜃→0sin 𝜃 (1 − cos 𝜃_𝜃cos 𝜃 )₃

= lim_𝜃→0sin 𝜃 (2 𝑠𝑖𝑛 21

(41)

SMAN 12 MAKASSAR

= 2lim_𝜃→0tan 𝜃 ( 𝑠𝑖𝑛 21

2 𝜃) 𝜃3

= 2lim_𝜃→0tan 𝜃_{𝜃 . (lim}_𝜃→0sin 12𝜃_{𝜃 )} 2

= 2.1. (1 2)

2

=2.1 4 =1₂ Jawaban B

101. Jika lim_𝑥→0_{tan 𝑥−sin 𝑥}𝑥3 = 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 = lim𝑥→0

𝑥3

tan 𝑥 (1 − 𝑐𝑜𝑠𝑥) = lim_𝑥→0 𝑥3

tan 𝑥 . 2 𝑠𝑖𝑛21 2 𝑥 =1_{2 lim}_𝑥→0_{tan 𝑥 . lim}𝑥 _𝑥→0 𝑥2

𝑠𝑖𝑛21 2 𝑥

=1_{2 lim}_𝑥→0_{tan 𝑥 . (lim}𝑥 _𝑥→0 𝑥 𝑠𝑖𝑛 12𝑥)

2

=1_{2 . 1.}1₁ 4 = 2

Nilai dari 𝐴 − 2 = 2 sehingga A = 4 Jadi A+2 = 4 + 2 = 6

Jawaban E

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

(42)

SMAN 12 MAKASSAR

= lim_𝑥→2sin(𝑥 − 2)_{3(𝑥 − 2) . lim}_𝑥→2sin(𝑥 − 2)_{(𝑥 − 2)}

= 1_{3 . 1}

= 1₃

Jawaban A

103. Nilai lim 𝑥→𝜋₄

(𝑥−𝜋₄) sin(3𝑥−3𝜋₄) 2(1−sin 2𝑥) = ⋯.

A. 3

4 D.−

3 4 B. 1

4 E. −

3 2 C. 0

Pembahasan

Misalkan 𝑥 −𝜋₄ = 𝑦 ⇔ 𝑥 = 𝑦 +𝜋₄ sehingga 2𝑥 = 2𝑦 +𝜋₂

Jika 𝑥 →𝜋

4 maka 𝑦 → 0

lim 𝑥→𝜋4

(𝑥 − 𝜋4)sin(3𝑥 −3𝜋4 )

2(1 − sin 2𝑥) = lim_𝑥→𝜋4(𝑥 − 𝜋4)sin3(𝑥 − 𝜋 4) 2(1 − sin 2𝑥) = lim_𝑦→0 𝑦 sin 3𝑦

2 (1 − sin (2𝑦 + 𝜋2)) = lim_𝑦→0_{2. (1 − 𝑐𝑜𝑠2𝑦)}𝑦 sin 3𝑦

=1 2 lim𝑦→0

𝑦 sin 3𝑦 2. sin2_y

=1_{4 . lim}_𝑦→0_{sin 𝑦 . lim}𝑦 _𝑦→0sin 3𝑦_{sin 𝑦}

=1_{4 . 1.3}

=3₄ Jawaban A

104. Nilai lim 𝑥→𝜋₂

4(𝑥−𝜋)cos2x

𝜋(𝜋−2𝑥) tan(𝑥−𝜋₂) = ⋯.

A. −2 D. 1

B. −1 E. 2

C. 0

Pembahasan

Misalkan 𝑦 = 𝑥 −𝜋₂ maka 𝑥 =𝜋₂+ 𝑦

lim 𝑥→𝜋2

4(𝑥 − 𝜋)cos2_x

𝜋(𝜋 − 2𝑥) tan (𝑥 − 𝜋2) = lim𝑦→0

(43)

SMAN 12 MAKASSAR

= lim_𝑦→04 (𝑦 − 𝜋2)cos 2

(𝜋2 + 𝑦) 𝜋(−𝑦) tan 𝑦 = lim_𝑦→0(4𝑦 − 2𝜋)(− sin 𝑦)_{𝜋(−𝑦) tan 𝑦} 2

= lim_𝑦→0(4𝑦 − 2𝜋)_−𝜋 . lim_𝑦→0sin 𝑦_{𝑦 . lim}_𝑦→0_{tan 𝑦}sin 𝑦

=(4.0 − 2𝜋)_−𝜋 . 1.1₁

=−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 − 𝑥𝑦)(1 + tan𝑥 tan𝑦) = lim𝑥→𝑦

tan(𝑥 − 𝑦) (1 − 𝑥𝑦) = lim_𝑥→𝑦tan(𝑥 − 𝑦)

(𝑦 − 𝑥_{𝑦 )}

= lim_𝑥→𝑦tan(𝑥 − 𝑦)₁ 𝑦 (𝑦 − 𝑥) = 𝑦lim_𝑥→𝑦tan(𝑥 − 𝑦)_{(𝑦 − 𝑥)}

= 𝑦lim_𝑥→𝑦tan(𝑥 − 𝑦) −(𝑥 − 𝑦) = −𝑦lim_𝑥→𝑦tan(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 − 𝑎𝑏)tan𝑎 tan𝑏 −𝑎𝑏

= lim_𝑎→𝑏 tan 𝑎 − tan 𝑏

(44)

SMAN 12 MAKASSAR

= lim_𝑎→𝑏 tan 𝑎 − tan 𝑏 (1 − 𝑎𝑏)(1 + tan𝑎 tan𝑏) = lim_𝑎→𝑏tan(𝑎 − 𝑏)

(1 − 𝑎𝑏) = lim_𝑎→𝑏tan(𝑎 − 𝑏)

(𝑏 − 𝑎_{𝑏 )}

= lim_𝑎→𝑏tan(𝑎 − 𝑏)₁ 𝑏 (𝑏 − 𝑎) = 𝑏lim_𝑎→𝑏tan(𝑎 − 𝑏)_{(𝑏 − 𝑎)}

= 𝑏lim_𝑎→𝑏tan(𝑎 − 𝑏)_{−(𝑎 − 𝑏)}

= −𝑏lim_𝑎→𝑏tan(𝑎 − 𝑏)_{(𝑎 − 𝑏)} = −𝑏. 1

= −𝑏 Jawaban E

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 → 𝜋₂ maka 𝑦 → 0

lim 𝑥→𝜋2

2𝑥 − 𝜋

cos 𝑥 = lim𝑦→0

𝑦 cos (𝜋2 +𝑦2)

= lim_𝑦→0 𝑦 −sin 𝑦2 =

1

− 12= −2 Jawaban D

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 = lim𝑦→0

sin 𝜋(𝑦 + 1) 𝑦 = lim𝑦→0

sin(𝜋𝑦 + 𝜋) 𝑦 = lim𝑦→0

−sin 𝜋𝑦

𝑦 = −𝜋

(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 = lim𝑦→0

tan 𝜋(𝑦 − 2) 𝑦 = lim𝑦→0

tan(𝜋𝑦 − 2𝜋) 𝑦 = lim𝑦→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→01 − cos 𝑦_𝑦₂

= lim_𝑦→02 𝑠𝑖𝑛 21

2 𝑦 𝑦2

= 2 (lim_𝑦→0sin 12𝑦_{𝑦 )} 2

= 2.1₂2

=2₄ = 0,50

Jawaban E

sin 2𝑥 𝑥− 𝜋₂ =...

A. −2 D. 1

B. −1 E. 2

(46)

SMAN 12 MAKASSAR

Pembahasan

Misalkan 𝑥 − 𝜋₂ = 𝑦 atau 𝑥 =𝜋₂+ 𝑦

sin 2𝑥 = sin 2 (𝜋_{2 + 𝑦) = sin}(𝜋 + 𝑦) = − sin 𝑦

lim 𝑥→𝜋2

sin 2𝑥 𝑥 − 𝜋2 = lim𝑦→0

− sin 𝑦

𝑦 = − lim𝑦→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 = lim𝑦→0 𝑦

−sin 𝑦 = − lim𝑦→0 𝑦

sin 𝑦 = −1 Jawaban B

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}𝑦₂

Jika 𝑥 → 𝜋₂ maka 𝑦 → 0

lim 𝑥→𝜋2

1 − sin 𝑥

(𝜋 − 2𝑥)2 _{= lim}

𝑦→01 − cos 𝑦2𝑦2

= lim_𝑦→02 𝑠𝑖𝑛 2𝑦

4 𝑦2

= 2lim_𝑦→0sin 𝑦4_{𝑦 . lim}_𝑦→0sin 𝑦4_𝑦

= 2.1_{4 .}1₄

=1₈

(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→0𝑦 tan (𝜋(1 − 𝑦)₂ ) = lim_𝑦→0𝑦 tan (𝜋

2 − 𝜋 2 𝑦) = lim_𝑦→0𝑦 cot (𝜋_{2 𝑦)} = lim_𝑦→0 𝑦

tan (𝜋2 𝑦) = 1_𝜋

2 = 2_𝜋 Jawaban B

115. Nilai lim 𝑥→𝜋

1+cos 𝑥 𝑡𝑎𝑛2_𝑥 =….

A. −1 D. 1₂

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→01 + cos(𝜋 + 𝑡)_𝑡𝑎𝑛2_{(𝜋 + 𝑡)} = lim_𝑡→01 − cos 𝑡_𝑡𝑎𝑛₂_𝑡

= lim_𝑡→02 𝑠𝑖𝑛 21

2 𝑡 𝑡𝑎𝑛2_𝑡

= 2 lim_𝑡→0sin 12𝑡_{tan 𝑡 . lim}_𝑡→0sin 12𝑡_{tan 𝑡}

= 2.1 2 .

1 2 =1₂

(48)

SMAN 12 MAKASSAR

116. lim_𝑥→1tan(𝑥_(𝑥−1)2−1)=….

A. 2 D. −2

B. 1₂ E. −1₂

C. 0

Pembahasan

lim 𝑥→1

tan(𝑥2 _{− 1)}

(𝑥 − 1) = lim𝑥→1

tan(𝑥2_{− 1)} (𝑥 − 1)

= lim_𝑥→1tan(𝑥_{(𝑥 − 1) .}2− 1) (𝑥 + 1)_{(𝑥 + 1)}

= lim_𝑥→1tan(𝑥_(𝑥₂2_{− 1) . lim}− 1) _𝑥→1(𝑥 + 1) = 1. (1 + 1)

= 2 Jawaban A

√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 = lim𝑥→0

√1 − cos 𝑥 √𝑥2

= lim_𝑥→0√1 − cos 𝑥 𝑥2

= √lim_𝑥→01 − cos 𝑥_𝑥₂

= √lim_𝑥→02 𝑠𝑖𝑛 21

2 𝑥 𝑥2

= √2lim_𝑥→0sin 12𝑥_{𝑥 . lim}_𝑥→0sin 12𝑥_𝑥

= √2.1_{2 .}1₂

=1_{2 √2}

(49)

SMAN 12 MAKASSAR

118. Jika diketahui lim 𝑥→0

𝑎𝑥 sin 𝑥+𝑏

cos 𝑥−1 = 1, maka nilai 𝑎 dan 𝑏yang memenuhi adalah …. A. 𝑎 = −1₂, 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_{cos 𝑥 − 1 = 1}𝑎𝑥 sin 𝑥

⇔ lim_𝑥→0 𝑎𝑥 sin 𝑥 −2 𝑠𝑖𝑛21

2 𝑥 = 1

⇔ lim_𝑥→0 𝑎𝑥 sin 12𝑥. lim𝑥→0

sin 𝑥

sin 12𝑥 = −2 ⇔ 𝑎₁

2 .1₁

2 = −2

⇔ 𝑎 = −2 ×1_{4 = −}1₂ 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. 𝛼₂2(𝛼 + 𝛽)2

B. 1₂(𝛼 − 𝛽)2 _E.𝛽2

2 (𝛼 − 𝛽)2 C. 𝛼₂2(𝛼 − 𝛽)2

Pembahasan

lim 𝑥→𝛼

1 − cos(𝑎𝑥2_{+ 𝑏𝑥 + 𝑐)}

(𝑥 − 𝛼)2 _{= lim} 𝑥→𝛼

2 sin21

2 (𝑎𝑥2+ 𝑏𝑥 + 𝑐) (𝑥 − 𝛼)2

= lim_𝑥→𝛼2 sin 21

2 (𝑥 − 𝛼)(𝑥 − 𝛽) (𝑥 − 𝛼)2

= 2 (lim

𝑥→𝛼sin 12(𝑥 − 𝛼)(𝑥 − 𝛽)(𝑥 − 𝛼) ) 2

= 2 (lim_𝑥→𝛼sin 12(𝑥 − 𝛼)(𝑥 − 𝛽)_{(𝑥 − 𝛼)} .(𝑥 − 𝛽)_{(𝑥 − 𝛽))} 2

(50)

SMAN 12 MAKASSAR

= 2 (lim

𝑥→𝛼(𝑥 − 𝛽)lim𝑥→𝛼sin 12(𝑥 − 𝛼)(𝑥 − 𝛽)(𝑥 − 𝛼)(𝑥 − 𝛽) ) 2

= 2 ((𝛼 − 𝛽).1₂₎ 2

= 2.1₄(𝛼 − 𝛽)2

=1₂(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 cos1₂(2𝑥 + 4ℎ + 2𝑥 − 4ℎ) cos1₂(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→02 cos 2𝑥 (cos 4ℎ − 1)_4ℎ₂

= lim_ℎ→02 cos 2𝑥 (−2. 𝑠𝑖𝑛_{4. ℎ}₂ 22ℎ)

= lim_ℎ→0− cos 2𝑥 . sin _ℎ₂ 22ℎ

= lim_ℎ→0− cos 2𝑥 lim_ℎ→0sin _ℎ2₂2ℎ

= − cos 2𝑥 lim_ℎ→0sin 2ℎ_{ℎ . lim}_ℎ→0sin 2ℎ_ℎ = − cos 2𝑥 . 2.2

= −4 cos 2𝑥

Jawaban D