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Hasil dari x→0lim​2cos4x−25xtan2x​ adalah ...

Pertanyaan

Hasil dari undefined adalah ...

  1. size 14px 5 over size 14px 8    

  2. size 14px 8 over size 14px 5    

  3. size 14px 1    

  4. size 14px minus size 14px 8 over size 14px 5     

  5. size 14px minus size 14px 5 over size 14px 8   

N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan bahwa

begin mathsize 14px style cos invisible function application alpha equals 1 minus 2 sin squared invisible function application open parentheses 1 half alpha close parentheses end style 

Jika disubstitusi begin mathsize 14px style alpha equals 4 x end style, maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application 4 x end cell equals cell 1 minus 2 sin squared invisible function application open parentheses 1 half times 4 x close parentheses end cell row blank equals cell 1 minus 2 sin squared invisible function application 2 x end cell end table end style          

Oleh karena itu, didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 cos invisible function application 4 x minus 2 end cell equals cell 2 open parentheses 1 minus 2 sin squared invisible function application 2 x close parentheses minus 2 end cell row blank equals cell 2 minus 4 sin squared invisible function application 2 x minus 2 end cell row blank equals cell negative 4 sin squared invisible function application 2 x end cell end table end style             

Kemudian, didapat bahwa

begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator 2 cos invisible function application 4 x minus 2 end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator negative 4 sin squared invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator negative 4 sin invisible function application 2 x sin invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x over denominator negative 4 sin invisible function application 2 x end fraction times limit as x rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 2 x over denominator sin invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator 5 x over denominator negative 4 sin invisible function application 2 x end fraction times 2 over 2 close parentheses times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 2 x over denominator sin invisible function application 2 x end fraction times fraction numerator 2 x over denominator 2 x end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 x over denominator sin invisible function application 2 x end fraction times fraction numerator 5 over denominator negative 4 times 2 end fraction close parentheses times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 2 x over denominator 2 x end fraction times fraction numerator 2 x over denominator sin invisible function application 2 x end fraction close parentheses equals open parentheses 1 times fraction numerator 5 over denominator negative 8 end fraction close parentheses times open parentheses 1 times 1 close parentheses equals negative 5 over 8 end style          

Jadi, jawaban yang tepat adalah E.

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