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Pertanyaan

x → 0 lim ​ sin 2 2 x 1 − cos 2 x ​ = ...

                   

  1. 0 comma 125      

  2. 0 comma 25        

  3. 0 comma 5 

  4. table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table           

  5.  infinity   

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P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

jawaban yang benar adalah C.

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Pembahasan

Ingatlah identitas cosinus sudut rangkap serta sifat-sifat limit fungsi Dengan menggunakan identitas sudut rangkap dan sifat-sifat limit fungsi tersebut,didapatkan: Maka, . Oleh karena itu, jawaban yang benar adalah C.

Ingatlah identitas cosinus sudut rangkap

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

serta sifat-sifat limit fungsi

  • limit as x rightwards arrow c of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow c of f left parenthesis x right parenthesis
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis close square brackets to the power of n equals open square brackets limit as x rightwards arrow c of f left parenthesis x right parenthesis close square brackets to the power of n
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b

Dengan menggunakan identitas sudut rangkap 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 space 2 x over denominator sin squared space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator 1 minus open parentheses 1 minus 2 sin squared space x close parentheses over denominator sin squared space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator 2 sin squared space x over denominator sin squared space 2 x end fraction end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space fraction numerator sin squared space x over denominator sin squared space 2 x end fraction end cell row blank equals cell 2 times open parentheses limit as x rightwards arrow 0 of space fraction numerator sin space x over denominator sin space 2 x end fraction close parentheses squared end cell row blank equals cell 2 times open parentheses 1 half close parentheses squared end cell row blank equals cell 2 times open parentheses 1 fourth close parentheses end cell row blank equals cell 1 half end cell row blank equals cell 0 comma 5 end cell end table            

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

Oleh karena itu, jawaban yang benar adalah C.

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