Nilai x→0lim sin4x+sin7xtan3x adalah ....

Pertanyaan

Nilai $limit as x rightwards arrow 0 of space fraction numerator tan 3 x over denominator sin 4 x plus sin 7 x end fraction$ adalah ....

$\frac2{11}$
$\frac3{11}$
$\frac4{11}$
$\frac7{11}$
$\frac{12}{11}$

W. Rohmiyati

Master Teacher

Mahasiswa/Alumni Universitas Langlangbuana

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Untuk mengerjakan soal limit fungsi trigonometri, kita substitusi terlebih dahulu menjadi:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator tan 3 x over denominator sin 4 x plus sin 7 x end fraction end cell equals cell fraction numerator tan 3 left parenthesis 0 right parenthesis over denominator sin 4 left parenthesis 0 right parenthesis plus sin 7 left parenthesis 0 right parenthesis end fraction end cell row blank equals cell fraction numerator tan 0 over denominator sin 0 plus sin 0 end fraction end cell row blank equals cell fraction numerator 0 over denominator 0 plus 0 end fraction end cell row blank equals cell 0 over 0 space left parenthesis Nilai space tak space tentu right parenthesis space end cell end table$

Karena diperoleh hasil bentuk tak tentu, maka limit fungsi trigonometri harus diubah lalu gunakan konsep berikut ini:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction end cell equals cell a over b end cell row cell limit as x rightwards arrow 0 of space fraction numerator tan space a x over denominator b x end fraction end cell equals cell a over b end cell end table$

Maka soal di atas, dikerjakan dengan cara sebagai berikut :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator tan 3 x over denominator sin 4 x plus sin 7 x end fraction cross times fraction numerator begin display style 1 over x end style over denominator begin display style 1 over x end style end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator begin display style fraction numerator tan 3 x over denominator x end fraction end style over denominator begin display style fraction numerator sin 4 x over denominator x end fraction end style plus begin display style fraction numerator sin 7 x over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator 3 over denominator 4 plus 7 end fraction end cell row blank equals cell 3 over 11 end cell end table$

Oleh karena itu, jawaban yang benar adalah B.