Pertanyaan

x → 0 lim sin 2 x − sin x sin 3 x + sin x = ...

$x \to 0 lim \frac{sin 3 x + sin x}{sin 2 x - sin x} = ...$

P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Ingatlah sifat-sifat limit fungsi: Dengan menggunakan sifat-sifat limit fungsi tersebut, didapatkan : Maka, . Oleh karena itu, jawaban yang benar adalah A.

Ingatlah sifat-sifat limit fungsi:

$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals a over b$

Dengan menggunakan sifat-sifat limit fungsi tersebut, didapatkan :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator sin space 3 x plus sin space x over denominator sin space 2 x minus sin space x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses sin space 3 x plus sin space x close parentheses over denominator open parentheses sin space 2 x minus sin space x close parentheses end fraction cross times fraction numerator 1 over x over denominator 1 over x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space open parentheses fraction numerator begin display style fraction numerator sin space 3 x over denominator x end fraction plus fraction numerator sin space x over denominator x end fraction end style over denominator begin display style fraction numerator sin space 2 x over denominator x end fraction minus fraction numerator sin space x over denominator x end fraction end style end fraction close parentheses end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space 3 x over denominator x end fraction plus fraction numerator sin space x over denominator x end fraction close parentheses end style over denominator begin display style limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space 2 x over denominator x end fraction minus fraction numerator sin space x over denominator x end fraction close parentheses end style end fraction end cell row blank equals cell fraction numerator limit as x rightwards arrow 0 of space begin display style fraction numerator sin space 3 x over denominator x end fraction end style plus limit as x rightwards arrow 0 of space begin display style fraction numerator sin space x over denominator x end fraction end style over denominator limit as x rightwards arrow 0 of space begin display style fraction numerator sin space 2 x over denominator x end fraction end style minus limit as x rightwards arrow 0 of begin display style fraction numerator sin space x over denominator x end fraction end style space end fraction end cell row blank equals cell fraction numerator 3 plus 1 over denominator 2 minus 1 end fraction end cell row blank equals cell 4 over 1 end cell row blank equals 4 end table$

Maka, $limit as x rightwards arrow 0 of space fraction numerator sin space 3 x plus sin space x over denominator sin space 2 x minus sin space x end fraction equals 1$ .

Oleh karena itu, jawaban yang benar adalah A.