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Pertanyaan

Hasil dari x → 0 lim ​ 48 x 3 sin 3 x − sin 3 x cos 6 x ​ adalah ...

Hasil dari  adalah ...

  1. size 14px minus size 14px 11 over size 14px 8    

  2. size 14px minus size 14px 9 over size 14px 8       

  3. size 14px minus size 14px 7 over size 14px 8       

  4. size 14px 7 over size 14px 8       

  5. size 14px 9 over size 14px 8         

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N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Ingat bahwa Oleh karena itu, didapat Jadi, jawaban yang tepat adalah E.

Ingat 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     

Oleh karena itu, didapat

begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 3 x minus sin invisible function application 3 x cos invisible function application 6 x over denominator 48 x cubed end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 3 x open parentheses 1 minus cos invisible function application 6 x close parentheses over denominator 3 x times 16 x squared end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 3 x over denominator 3 x end fraction times limit as x rightwards arrow 0 of invisible function application fraction numerator 1 minus cos invisible function application 6 x over denominator 16 x squared end fraction equals 1 times limit as x rightwards arrow 0 of invisible function application fraction numerator 1 minus open parentheses 1 minus 2 sin squared invisible function application open parentheses 1 half times 6 x close parentheses close parentheses over denominator 16 x squared end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 1 minus open parentheses 1 minus 2 sin squared invisible function application 3 x close parentheses over denominator 16 x squared end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 2 sin squared invisible function application 3 x over denominator 16 x squared end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin squared invisible function application 3 x over denominator 8 x squared end fraction equals 1 over 8 times limit as x rightwards arrow 0 of invisible function application fraction numerator sin squared invisible function application 3 x over denominator x squared end fraction equals 1 over 8 times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator sin invisible function application 3 x over denominator x end fraction close parentheses squared equals 1 over 8 times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator sin invisible function application 3 x over denominator x end fraction times 3 over 3 close parentheses squared equals 1 over 8 times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator sin invisible function application 3 x over denominator 3 x end fraction times 3 close parentheses squared equals 1 over 8 times open parentheses 1 times 3 close parentheses squared equals 1 over 8 times 9 equals 9 over 8 end style   

Jadi, jawaban yang tepat adalah E.

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3

Khouwsasi Morten

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Hasil dari x → 0 lim ​ sin 3 x cos 9 x − sin 3 x cos 3 x sin 6 x − tan 6 x ​ adalah ....

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