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Nilai dari

Pertanyaan

Nilai dari begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator sin 2 x cos 8 x minus sin 2 x cos 2 x over denominator 2 x cubed end fraction equals.... end stylespace 

  1. begin mathsize 14px style negative 30 end style 

  2. begin mathsize 14px style negative 15 end style 

  3. begin mathsize 14px style negative 1 end style 

  4. begin mathsize 14px style negative 2 end style 

  5. begin mathsize 14px style negative 5 end style 

Pembahasan Soal:

Salah satu teorema untuk limit trigonometri adalah

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

Oleh karena itu, limit trigonometri pada soal akan diubah sehingga teorema di atas dapat digunakan. Dengan menggunakan sifat distributif, diperoleh 

limit as x rightwards arrow 0 of fraction numerator sin 2 x cos 8 x minus sin 2 x cos 2 x over denominator 2 x cubed end fraction equals limit as x rightwards arrow 0 of fraction numerator sin 2 x open parentheses cos 8 x minus cos 2 x close parentheses over denominator 2 x cubed end fraction

Lalu, dengan menggunakan rumus selisih trigonometri, cos A minus cos B equals negative 2 sin 1 half open parentheses A plus B close parentheses sin 1 half open parentheses A minus B close parentheses, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x open parentheses negative 2 sin begin display style 1 half end style open parentheses 8 x plus 2 x close parentheses sin begin display style 1 half end style open parentheses 8 x minus 2 x close parentheses close parentheses over denominator 2 x cubed end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x open parentheses negative 2 sin begin display style 1 half end style open parentheses 10 x close parentheses sin begin display style 1 half end style open parentheses 6 x close parentheses close parentheses over denominator 2 x cubed end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x open parentheses negative 2 sin 5 x sin 3 x close parentheses over denominator 2 x cubed end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x times open parentheses negative 2 sin 5 x close parentheses times sin 3 x over denominator 2 x times x times x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x over denominator 2 x end fraction times limit as x rightwards arrow 0 of fraction numerator open parentheses negative 2 sin 5 x close parentheses over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin 3 x over denominator x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin 2 x over denominator 2 x end fraction times open parentheses negative 2 close parentheses limit as x rightwards arrow 0 of fraction numerator sin 5 x over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin 3 x over denominator x end fraction end cell row blank equals cell 2 over 2 times open parentheses negative 2 close parentheses times 5 over 1 times 3 over 1 end cell row blank equals cell 1 times open parentheses negative 2 close parentheses times 5 times 3 end cell row blank equals cell negative 30 end cell end table

Jadi, limit as x rightwards arrow 0 of fraction numerator sin 2 x cos 8 x minus sin 2 x cos 2 x over denominator 2 x cubed end fraction equals negative 30

Oleh karena itu, jawaban yang benar adalah A.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 27 April 2021

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