x → 0 lim ​ sin 2 2 1 ​ x tan 2 6 1 ​ x ​ = ....

Pembahasan

Jawaban yang benar dari pertanyaan tersebut adalah C. Perhatikan konsep : tan 2 x = ( tan x ) 2 dan sin 2 x = ( sin x ) 2 . Maka soal dapat diselesaikan seperti berikut : lim x → 0 ​ s i n 2 2 1 ​ x t a n 2 6 1 ​ x ​ ​ = = = = = = ​ lim x → 0 ​ ( s i n 2 1 ​ x ) 2 ( t a n 6 1 ​ x ) 2 ​ ← Ingat tan x = c o s x s i n x ​ lim x → 0 ​ ( s i n 2 1 ​ x ) 2 ( c o s 6 1 ​ x s i n 6 1 ​ x ​ ) ​ 2 lim x → 0 ​ ( s i n 2 1 ​ x ) 2 ( cos 6 1 ​ x ) 2 ( sin 6 1 ​ x ) 2 ​ ​ lim x → 0 ​ ( c o s 6 1 ​ x ) 2 ( s i n 6 1 ​ x ) 2 ​ ⋅ ( s i n 2 1 ​ x ) 2 1 ​ lim x → 0 ​ c o s 6 1 ​ x ⋅ c o s 6 1 ​ x ⋅ s i n 2 1 ​ x ⋅ s i n 2 1 ​ x s i n 6 1 ​ x ⋅ s i n 6 1 ​ x ​ lim x → 0 ​ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ c o s 6 1 ​ x ⋅ c o s 6 1 ​ x 1 ​ ​ Ingat sifat x → 0 lim ​ f ( x ) ⋅ g ( x ) = x → 0 lim ​ f ( x ) ⋅ x → 0 lim ​ g ( x ) dan x → 0 lim ​ sin b x sin a x ​ = b a ​ , sehingga diperoleh : lim x → 0 ​ s i n 2 2 1 ​ x t a n 2 6 1 ​ x ​ ​ = = = = = ​ lim x → 0 ​ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ c o s 6 1 ​ x ⋅ c o s 6 1 ​ x 1 ​ lim x → 0 ​ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ lim x → 0 ​ s i n 2 1 ​ x s i n 6 1 ​ x ​ ⋅ x → 0 lim ​ c o s 6 1 ​ x ⋅ c o s 6 1 ​ x 1 ​ 2 1 ​ 6 1 ​ ​ ⋅ 2 1 ​ 6 1 ​ ​ ⋅ c o s 0 ⋅ c o s 0 1 ​ 3 1 ​ ⋅ 3 1 ​ ⋅ 1 ⋅ 1 1 ​ 9 1 ​ ​ Oleh karena itu, jawaban yang benar adalah C.