Hasil x→0lim​2x⋅sin 3x1−cos2 2x​ adalah ....

Pertanyaan

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

  1. 4 over 3

  2. 2 over 3

  3. 1 third

  4. 3 over 4

  5. 3 over 2

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah B.

Soal tersebut merupakan soal limit fungsi trigonometri yang memuat sinus dan cosinus. Jika diselesaikan dengan metode substitusi, maka:

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

Karena 0 over 0 tidak tentu, maka soal tersebut perlu diselesaikan dengan menggunakan cara lain.

Ingat bahwa terdapat identitas trigonometri:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared space alpha plus cos squared space alpha end cell equals 1 row cell sin squared space alpha end cell equals cell 1 minus cos squared space alpha end cell end table

Pada limit fungsi trigonometri yang memuat sinus, berlaku:

limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator a x end fraction equals 1  limit as x rightwards arrow 0 of fraction numerator a x over denominator sin space a x end fraction equals 1

Diperoleh,

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

Dengan demikian, hasil limit as x rightwards arrow 0 of fraction numerator 1 minus cos squared space 2 x over denominator 2 x times sin space 3 x end fraction adalah 2 over 3.

Oleh karena itu, jawaban yang benar adalah B.

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