9. Tentukan nilai limt berikut
lim
π₯β0
cos 7π₯ β cos 5π₯
1 β cos 3π₯
Jawab :
lim
π₯β0
β2 sin
1
2
7π₯+5π₯ sin
1
2
(7π₯β5π₯)
1β(1β2 sin 2 3
2
π₯)
=
lim
π₯β0
β2 sin
1
2
7π₯+5π₯ sin
1
2
(7π₯β5π₯)
2 sin 2 3
2
π₯
=
lim
π₯β0
β2 sin 6π₯ sin π₯
2 sin
3
2
π₯ sin
3
2
π₯
=
β2 x 6 x 1
2 x
3
2
x
3
2
=
β12
9
2
= -
8
3 alewoh.com
10. Tentukan nilai limit berikut
lim
π₯β0
sin 7π₯ βtan 5π₯
tan 4π₯ β3π₯
Jawab :
lim
π₯β0
sin 7π₯ βtan 5π₯
tan 4π₯ β3π₯
lim
π₯β0
7.β5
4β3
=
2
1
= 2
alewoh.com
11. Tentukan nilai limit berikut
lim
π₯βπ
tan 4π₯ β sin 4π₯
π₯ sin 2 4π₯
Jawab :
lim
π₯βπ
tan 4π₯βsin 4π₯
π₯ sin 2
4π₯
= lim
π₯βπ
sin 4π₯βsin 4π₯ x cos 4π₯
cos 4π₯
π₯ sin 2
4π₯
= lim
π₯βπ
sin 4π₯ (1βcos 4π₯)
cos 4π₯
π₯ sin 2
4π₯
= lim
π₯βπ
tan 4π₯ (2 sin 2
2π₯)
π₯ sin 2
4π₯
= 2 lim
π₯βπ
tan 4π₯
π₯
lim
π₯βπ
sin 2π₯
sin 4π₯
lim
π₯βπ
sin 2π₯
sin 4π₯
= 2 x 4 x
2
4
x
2
4
= 2
alewoh.com