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Limit Tak Hingga Trigonometri
1.
lim
π₯ββ
A. B.
2 π₯ 6 sin π₯
sin
Pembahasan:
= β¦.
1
1
D. 3
6 1
E. 6
3
C. 2 Pembahasan: Misal π₯ =
1 π
βπ=
1 3 π₯ 2 π₯ββ 1βcos( ) π₯
5. Jika lim
7 3 5 sin +tan βsin π₯ π₯ π₯ 9 3 1 π₯ββ tanπ₯βtanπ₯βsinπ₯
lim
= β¦.
A. 1 B. 3 C. 5
D. 7 E. 9
Pembahasan: 1
πβ0
3.
sin 7π + tan 3π β sin 5π 7 + 3 β 5 = =1 tan 9π β tan 3π β sin π 9β3β1 2 π₯ 1 π₯ββ tan3 ( ) 2π₯
sin3 ( )
lim
= β¦.
A. 23 B. 24 C. 25
D. 26 E. 27
B. 3
E. 2
6.
1
1 4
2
3
sin 2π 2 lim = ( ) = 43 = 26 1 1 πβ0 tan3 (2 π) 2 2 π₯
3 π₯
1
3
A. 0 2 3 3 2
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adalah β¦.
D. 2 E. 6
2
A. β 3
D. 3
2
E. 1
3
1 3
Pembahasan: Misal π₯ =
1 π
βπ=
lim (csc 2 2π β
πβ0
π₯ 2 tan( ) tan( )
1
lim [csc 2 (π₯ ) β 4 π₯ 2 ] = β¦.
π₯ββ
C.
3
C.
, maka nilai π adalah β¦.
tan2 3π π2 = πβ0 1 β cos 2π 2 tan2 3π π2 lim = πβ0 2 sin2 π 2 9 π2 = β π = Β±3 2 2
Misal π₯ = π β π = π₯
B.
2
Pembahasan: 1 1 Misal π₯ = π β π = π₯
1
π₯ββ
π2
D. 1
C.
Pembahasan:
4. Nilai dari lim
=
A. 4
B. β 1
tan2( )
lim
1
Misal π₯ = π β π = π₯ lim
1 ) tan 2π tan 3π tan 2π tan 3π π2 lim = lim ( Γ ) πβ0 πβ0 3 3π π 2 = Γ3 3 =2 (
π₯
sin 2π 2 1 lim = = πβ0 sin 6π 6 3 2.
1
Misal π₯ = π β π = π₯
1 π₯
1 1 1 ) = lim ( 2 β 2) πβ0 sin 2π 4π 4π 2 2 4π β sin 2π = lim ( ) πβ0 4π2 . sin2 2π
Dengan menggunakan nilai hampiran (maclaurin) diperoleh:
1
Limit Tak Hingga Trigonometri 4π2 β sin2 2π lim ( ) πβ0 4π2 . sin2 2π 8π3 4π β (2π β 6 ) = lim πβ0 4π2 . (2π )2
1
C. β 2
2
2
Pembahasan: 1
(
) 32 4 64 6 2 2 4π β (4π β 6 π + 36 π ) = lim ( ) πβ0 16π4 32 4 64 6 π β 36 π = lim ( 6 ) πβ0 16π4 32 64 2 β π = lim ( 6 36 ) πβ0 16 32 = 6 16 1 = 3
sin2 π β cos π + 1 sin2 π β (cos π β 1) = lim πβ0 πβ0 π tan π π tan π 1 sin2 π + 2 sin2 (2 π) = lim πβ0 π tan π 1 π2 + 2 ( π2 ) 4 = 2 π 1 =1+ 2 3 = 2 lim
9. SBMPTN 2017 SAINTEK 124 lim π₯ (sec
π₯ββ
7. Nilai dari lim
π₯ββ
2 π₯ 2 tan( ) π₯
4π₯.sin2( )
D. 4 E. 8
4 ( ) sin2 2π 4 sin2 2π π lim = lim πβ0 πβ0 π tan 2π tan 2π 4 sin 2π sin 2π = lim ( Γ ) πβ0 π tan 2π = 4(2)(1) =8 lim
π₯ββ
A. B.
E. β1
2
C. 0
1 lim (sec βπ β 1) πβ0 π Misal βπ = π lim
= lim
3 2 1 2
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(
1
1
β 1) = lim π 2 ( πβ0
1βcos π cos π
)
1 2 sin2(2 π)
πβ0
π 2 cos π
1 2. 4 π 2
1 = 2
= β¦.
1
πβ0 π 2 cos π
= 1 1 π₯ π₯ 1 1 ( ) tan( ) π₯ π₯
1
D. β 2
Pembahasan: 1 1 Misal π₯ = π β π = π₯
π₯
sin2( )βcos( )+1
β 1) = β¦.
1
B.
Pembahasan: 1 1 Misal π₯ = β π = π
1 βπ₯
A. 1
adalah β¦.
A. β8 B. β4 C. 0
8.
1
Misal π₯ = π β π = π₯
π2
D. β1 E. β2
2
Limit Tak Hingga Trigonometri 10. SBMPTN 2017 SAINTEK 129 1 1 1 2π₯ 2 tan (π₯ ) β π₯ sin (π₯ ) + π₯ lim = β¦. 2 π₯ββ π₯ cos ( ) π₯ A. 2 D. β1 B. 1 E. β2 C. 0
π₯
π₯
β
1
) = ....
π₯+1
D. 1
2
1
E. 2
3 2
1
1
Misal π₯ = π β π₯ = π 1 1 1 lim (π₯ 3 sin ( ) + π₯) . ( β ) π₯ββ π₯ π₯β1 π₯+1
D. 1 E. +β
1
Misal π₯ = π β π = π₯
1 β cos π ) πβ0 sin π 1 2 sin2 2 π = lim ( ) πβ0 sin π
lim (csc π β cot π) = lim (
πβ0
1 π₯β1
Pembahasan:
Pembahasan: 1
1
π₯
lim (π₯ 3 sin ( ) + π₯) . (
C.
lim csc β cot = β¦.
A. ββ B. β1 C. 0
13. SBMPTN SAINTEK 2017 Kode 138
B. 2
11. SBMPTN 2017 SAINTEK 131 π₯ββ
1
1 cos π sin π2 lim ( ) cot π sin π2 = lim =1 πβ0 π πβ0 π sin π
A.
2 sin π tan π β +π π π π2 lim Γ cos 2π πβ0 π π 2 tan π 2 π β sin π + π = lim =2 πβ0 cos π
1
1
Misal π₯ = π β π = π₯
π₯ββ 5
Pembahasan: 1 1 Misal π₯ = π β π = π₯
1
Pembahasan:
1 2 sin 2 π 1 = lim ( ) sin π πβ0 sin π 2 =0
1 3 1 2π2 = lim (( ) sin π + ( )) ( ) πβ0 π π 1 β π2 1 sin π 2 )( + π) π2 ( ) 2 πβ0 π π 1 β π2 2 = lim (1 + π) ( ) πβ0 1 β π2 2 = (1 + 0) ( ) 1β0 =2 = lim (
14. SBMPTN 2017 SAINTEK 139 5 π₯ cot (π₯ + 1) lim = β¦. π₯ββ 1 β π₯2 1 A. β1 D. β 4 1
1
B. β 2
E. β 5
1
C. β 3 12. SBMPTN SAINTEK 2017 Kode 135/150 1 1 lim π₯ cot ( ) sin ( 2 ) = β¦. π₯ββ π₯ π₯ A. β2 D. 1 B. β1 E. 2 C. 0
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Pembahasan: 1
1
Misal π₯ = π β π = π₯ lim
π₯ββ
5 π₯ cot (π₯ + 1) 1 β π₯2
3
Limit Tak Hingga Trigonometri 1 π
( ) cot( 1
= 1lim π
5
+1 π 2 1
17. SBMPTN 2017 SAINTEK 146 2 4 csc 2 (π₯ ) β π₯ 3 sin (π₯ ) lim = β¦. π₯ββ π₯2 23 A. β 4
)
1β( )
ββ
π
1 5 ( ) cot( π+1 ) π π 1 1 πβ0 (1βπ)(1+π) 1 5π ( ) cot( ) π 1+π 1 1 πβ0 (π)(πβ1)(π)(π+1)
= lim
B. β C. β
= lim
D. β
1
= lim
1 π
5π ) 1+π
πβ0 (πβ1)( )(π+1) tan(
=β
E. β
1 5
21 4 19 4 17 4 15 4
Pembahasan: 15. SBMPTN SAINTEK 2017 KODE 140 3 sin π₯ lim = β¦. 2 1 π₯ββ (1 β cos ) . π₯ 2 sin π₯ π₯ 3 A. 0 D.
Misal π₯ =
B.
βπ=
csc 2 2π β lim
1 π₯
sin 4π π3
1 π2
πβ0
2
2
1 π
E. 3
3
C. 1 Pembahasan: 1
1
Misal π₯ = π β π = π₯ sin 3π
lim
πβ0
(1 β cos 2π) (
1 sin π) π2
sin 3π . π2 πβ0 2 sin2 π . sin π
= lim
3π 3 = 3 2π 3 = 2
16. SBMPTN 2017 SAINTEK 145/151 1 2 lim 2π₯ . tan . sec = β¦. π₯ββ π₯ π₯ A. 0 D. 3 B. 1 E. 4 C. 2 Pembahasan: 1
1 sin 4π β sin2 2π π3 = lim 1 πβ0 π2 π2 sin 4π = lim ( 2 β ) πβ0 sin 2π π 1 = β4 4 1 16 = β 4 4 15 =β 4
18. SBMPTN 2017 SAINTEK 147 2 cos2 (π₯ ) lim = β¦. 1 π₯ββ π₯ tan (2π₯ ) 1 A. 2 D. 4 B. 1 C. 2
E. 8
Pembahasan: 1
1
Misal π₯ = π β π = π₯
cos2 2π 1 lim = =2 1 1 πβ0 1 ( ) tan π π 2 2
1
Misal: π₯ = π β π = π₯
2 2 tan π lim ( ) tan π . sec 2π = lim =2 πβ0 π πβ0 π cos 2π
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Limit Tak Hingga Trigonometri 19. SBMPTN 2017 SAINTEK 149 3 sin π₯ lim = β¦. 4 π₯ββ (1 β cos ) π₯ π₯ 3 A. 0 D. 4 B. C.
2
E. 1
3 3
1
C.
1
E.
3 1
1 6
4
Pembahasan: 1
Misal π₯ = π β π = π₯ πβ0
B.
8
Pembahasan:
lim
20. SBMPTN 2017 SAINTEK 152 1 1 lim π₯ sec (1 β cos ) = β¦. π₯ββ π₯ βπ₯ 1 1 A. 2 D. 5
sin 3π 1 (1 β cos 4π) ( ) π
1
π sin 3π πβ0 2 sin2 2π
= lim
=
3 8
1
Misal π₯ = π β π = π₯
1 β cos βπ 1 lim ( ) sec π (1 β cos βπ) = lim ( ) πβ0 π πβ0 π cos π 1 2 sin2 2 βπ = lim πβ0 π 1 = 2( ) 4 1 = 2
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