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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.

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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