MATEMATIKA PEMINATAN
MATERI : LIMIT FUNGSI TRIGONOMETRI
MATERI : LIMIT FUNGSI TRIGONOMETRI
Nama : Sri Utami
Instansi : SMA Negeri 6 Surabaya
Instansi : SMA Negeri 6 Surabaya
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Gunakan metode substitusi untuk menentukan nilai Limit fungsi trigonometri berikut ini: 1. lim π₯β π 4 sin 2π₯ = sin 2 ( π 4 ) = sin 2π 4 = sin 900 = 1
2. lim π₯β 3π 4 tan 3π₯ + 2 = tan 3 ( 3π 4 ) + 2 = tan(450) + 2 = 1 + 2 = 3
3. limπ₯β0 sin π₯ sin π₯+cos π₯ = sin 0 sin 0+cos 0 = 0 0+1 = 0 4. lim π₯β π 2 1βcos 2π₯ 2 cos 2π₯ = 1βcos 2( π 2 ) 2 cos2( π 2 ) = 1βcosπ 2 cosπ = 1β(β1) 2 (β1) = 1+1 β2 = 2 β2 = β1
2. lim π₯β 3π 4 tan 3π₯ + 2 = tan 3 ( 3π 4 ) + 2 = tan(450) + 2 = 1 + 2 = 3
3. limπ₯β0 sin π₯ sin π₯+cos π₯ = sin 0 sin 0+cos 0 = 0 0+1 = 0 4. lim π₯β π 2 1βcos 2π₯ 2 cos 2π₯ = 1βcos 2( π 2 ) 2 cos2( π 2 ) = 1βcosπ 2 cosπ = 1β(β1) 2 (β1) = 1+1 β2 = 2 β2 = β1
video
https://www.youtube.com/watch?v=WbKFwA9DngI
Matematika hebat
https://www.youtube.com/watch?v=WbKFwA9DngI
Matematika hebat
Gunakan metode substitusi untuk menentukan nilai Limit fungsi trigonometri berikut ini: 1. lim π₯β π 4 sin 2π₯ = sin 2 ( π 4 ) = sin 2π 4 = sin 900 = 1
2. lim π₯β 3π 4 tan 3π₯ + 2 = tan 3 ( 3π 4 ) + 2 = tan(450) + 2 = 1 + 2 = 3
3. limπ₯β0 sin π₯ sin π₯+cos π₯ = sin 0 sin 0+cos 0 = 0 0+1 = 0 4. lim π₯β π 2 1βcos 2π₯ 2 cos 2π₯ = 1βcos 2( π 2 ) 2 cos2( π 2 ) = 1βcosπ 2 cosπ = 1β(β1) 2 (β1) = 1+1 β2 = 2 β2 = β1
2. lim π₯β 3π 4 tan 3π₯ + 2 = tan 3 ( 3π 4 ) + 2 = tan(450) + 2 = 1 + 2 = 3
3. limπ₯β0 sin π₯ sin π₯+cos π₯ = sin 0 sin 0+cos 0 = 0 0+1 = 0 4. lim π₯β π 2 1βcos 2π₯ 2 cos 2π₯ = 1βcos 2( π 2 ) 2 cos2( π 2 ) = 1βcosπ 2 cosπ = 1β(β1) 2 (β1) = 1+1 β2 = 2 β2 = β1
