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Pertanyaan

Diketahui: sin space A plus sin space B equals 2 space sin space 1 half open parentheses A plus B close parentheses times cos space 1 half open parentheses A minus B close parentheses dan cos space A minus cos space B equals negative 2 space sin space 1 half open parentheses A plus B close parentheses times sin space 1 half open parentheses A minus B close parentheses maka nilai dari limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction equals horizontal ellipsis 

  1. negative 5 over 3 

  2. negative 3 over 5 

  3. 0 

  4. 3 over 5 

  5. 5 over 3 

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Diketahui:

  • sin space A plus sin space B equals 2 space sin space 1 half open parentheses A plus B close parentheses times cos space 1 half open parentheses A minus B close parentheses
  • cos space A minus cos space B equals negative 2 space sin space 1 half open parentheses A plus B close parentheses times sin space 1 half open parentheses A minus B close parentheses

Ditanya:

  • nilai dari limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction ?

Penyelesaian:

Ingat kembali bahwa:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator sin space x over denominator x end fraction end cell equals 1 row cell limit as x rightwards arrow 0 of space fraction numerator sin space a x over denominator b x end fraction end cell equals cell a over b end cell end table

Misalkan: M equals limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction, maka nilai dari M adalah

table attributes columnalign right center left columnspacing 0px end attributes row M equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 1 half open parentheses 6 x plus 4 x close parentheses times cos space 1 half open parentheses 6 x minus 4 x close parentheses close parentheses over denominator negative 2 space sin space 1 half open parentheses 4 x plus 2 x close parentheses times sin space 1 half open parentheses 4 x minus 2 x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 1 half open parentheses 10 x close parentheses times cos space 1 half open parentheses 2 x close parentheses close parentheses over denominator negative 2 space sin space 1 half open parentheses 6 x close parentheses times sin space 1 half open parentheses 2 x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 5 x times cos space x close parentheses over denominator negative 2 space sin space 3 x times sin space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x times sin space 5 x times cos space x over denominator negative sin space 3 x times sin space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator begin display style x over x end style times begin display style fraction numerator sin space 5 x over denominator x end fraction end style over denominator negative begin display style fraction numerator sin space 3 x over denominator x end fraction end style times begin display style fraction numerator sin space x over denominator x end fraction end style end fraction times limit as x rightwards arrow 0 of cos space x end cell row blank equals cell fraction numerator 1 times 5 over denominator negative 3 times 1 end fraction times 1 end cell row M equals cell negative 5 over 3 end cell end table

Jadi, nilai dari limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction adalah negative 5 over 3. 

Oleh karena itu, jawaban yang benar adalah A.

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