x → 0 lim ​ sin 2 2 3 ​ x ( 1 − sec 2 2 x ) sin 4 3 x ​ = ...

Pembahasan

Untuk mengerjakan soal limit fungsi trigonometri, kita siubtitusi terlebih dahulu menjadi: lim x → 0 ​ s i n 2 2 3 ​ x ( 1 − s e c 2 2 x ) s i n 4 3 x ​ ​ = = = = = = ​ s i n 2 2 3 ​ ( 0 ) ( 1 − s e c 2 2 ( 0 ) s i n 4 3 ( 0 ) ​ s i n 2 ( 0 ) ( 1 − s e c 2 ( 0 )) s i n 4 ( 0 ) ​ ( s i n 0 ) 2 ( 1 − ( s e c 0 ) 2 ) ( s i n 0 ) 4 ​ 0 2 ( 1 − 1 2 ) 0 4 ​ 0 ( 0 ) 0 ​ 0 0 ​ ( Nilai tak tentu ) ​ Karena diperoleh bentuk tak tentu, maka limitfungsi trigonometri diubah lalu gunakan konsep x → 0 lim ​ sin b x sin a x ​ = b a ​ x → 0 lim ​ tan b x sin a x ​ = b a ​ serta identitas trigonometri berikut ini tan 2 x + 1 1 − sec 2 x ​ = = ​ sec 2 x − tan 2 x ​ Setelah itu, maka soal di atas diselesaikan dengan cara berikut ini: ​ = = = = = = = ​ lim x → 0 ​ s i n 2 2 3 ​ x ( 1 − s e c 2 2 x ) s i n 4 3 x ​ lim x → 0 ​ s i n 2 3 ​ x ⋅ s i n 2 3 ​ x ( − t a n 2 2 x ) s i n 3 x ⋅ s i n 3 x ⋅ s i n 3 x ⋅ s i n 3 x ​ lim x → 0 ​ s i n 2 3 ​ x ⋅ s i n 2 3 ​ x ⋅− t a n 2 x ⋅ t a n 2 x s i n 3 x ⋅ s i n 3 x ⋅ s i n 3 x ⋅ s i n 3 x ​ lim x → 0 ​ s i n 2 3 ​ x ​ ⋅ s i n 2 3 ​ x ​ ⋅− t a n 2 x ​ ⋅ t a n 2 x ​ s i n 3 x ​ ⋅ s i n 3 x ​ ⋅ s i n 3 x ​ ⋅ s i n 3 x ​ ​ 2 ​ 3 ​ ⋅ 2 ​ 3 ​ ⋅ ( − 2 ) ​ ⋅ 2 ​ 3 ⋅ 3 ⋅ 3 ⋅ 3 ​ 3 ⋅ 3 ⋅ ( − 1 ) 3 ⋅ 3 ⋅ 3 ⋅ 3 ​ − 1 3 ⋅ 3 ​ − 9 ​ Oleh karena itu, jawaban yang benar adalah E.