x→0lim x sin 2x1−cos2 3x= ...

Pertanyaan

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

P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

Pembahasan

Ingatlah identitas Trigonometri

1 minus cos squared space x equals sin squared space x

mengakibatkan

1 minus cos squared space 3 x equals sin squared space 3 x number space number space horizontal ellipsis left parenthesis 1 right parenthesis .

Ingat juga sifat-sifat limit fungsi

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

Dengan menggunakan bentuk (1) dan sifat-sifat limit fungsi tersebut, didapatkan:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator 1 minus cos squared space 3 x over denominator x space sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator sin squared space 3 x over denominator x space sin space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 3 x over denominator x end fraction times fraction numerator sin space 3 x over denominator sin space 2 x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin space 3 x over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space 3 x over denominator sin space 2 x end fraction end cell row blank equals cell 3 over 1 times 3 over 2 end cell row blank equals cell 9 over 2 end cell row blank equals cell 4 comma 5 end cell end table$

Maka, $limit as x rightwards arrow 0 of space fraction numerator 1 minus cos squared space 3 x over denominator x space sin space 2 x end fraction equals 4 comma 5$ .

Oleh karena itu, jawaban yang benar adalah D.