x→0lim​tan22xcos x−1​=...

Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator cos space x minus 1 over denominator tan squared 2 x end fraction equals... end style 

  1. undefined 

  2. undefined 

  3. undefined 

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R. Setiawan

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Pembahasan

Ingat bahwa:

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

Dan ingat rumus sudut rangkap untuk cosinus:

cos space x equals 1 minus 2 space sin squared space 1 half x 

Dari soal tersebut, dengan menggunakan sifat limit trigonometri dan rumus sudut rangkap untuk cosinus diperoleh hasil sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator cos space x minus 1 over denominator tan squared 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses 1 minus 2 sin squared begin display style space 1 half end style x close parentheses minus 1 over denominator tan squared 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator up diagonal strike 1 minus 2 sin squared begin display style space 1 half end style x minus up diagonal strike 1 over denominator tan squared 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator negative 2 sin squared begin display style 1 half end style x over denominator tan squared 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator negative 2 times sin space begin display style 1 half end style x times sin space begin display style 1 half end style x over denominator tan space 2 x times tan space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses negative 2 times fraction numerator sin space 1 half x over denominator tan space 2 x end fraction times fraction numerator sin space 1 half x over denominator tan space 2 x end fraction close parentheses end cell row blank equals cell negative 2 times fraction numerator 1 half over denominator 2 end fraction times fraction numerator 1 half over denominator 2 end fraction end cell row blank equals cell negative 1 over 8 end cell end table  

Oleh karena itu, jawaban yang benar adalah E.

70

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