Pertanyaan

Hasil dari adalah ...

Hasil dari $begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator tan squared invisible function application open parentheses 3 x minus 3 close parentheses over denominator sin invisible function application open parentheses 4 x minus 4 close parentheses squared end fraction end style$ adalah ...

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

M. Robo

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan bahwa Jadi, jawaban yang tepat adalah E.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator tan squared invisible function application open parentheses 3 x minus 3 close parentheses over denominator sin invisible function application open parentheses 4 x minus 4 close parentheses squared end fractionequals limit as x rightwards arrow 1 of invisible function application fraction numerator tan squared invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator sin invisible function application open parentheses 4 open parentheses x minus 1 close parentheses close parentheses squared end fractionequals limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses times tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator sin invisible function application open parentheses 16 open parentheses x minus 1 close parentheses squared close parentheses end fractionequals limit as x rightwards arrow 1 of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses times tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator sin invisible function application open parentheses 16 open parentheses x minus 1 close parentheses squared close parentheses end fraction times open parentheses x minus 1 close parentheses squared over open parentheses x minus 1 close parentheses squared close parenthesesequals limit as x rightwards arrow 1 of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x minus 1 close parentheses end fraction times fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x minus 1 close parentheses end fraction times fraction numerator open parentheses x minus 1 close parentheses squared over denominator sin invisible function application open parentheses 16 open parentheses x minus 1 close parentheses squared close parentheses end fraction close parenthesesequals limit as x rightwards arrow 1 of invisible function application open parentheses fraction numerator 3 times 3 over denominator 16 end fraction times fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator 3 open parentheses x minus 1 close parentheses end fraction times fraction numerator tan invisible function application open parentheses 3 open parentheses x minus 1 close parentheses close parentheses over denominator 3 open parentheses x minus 1 close parentheses end fraction times fraction numerator 16 open parentheses x minus 1 close parentheses squared over denominator sin invisible function application open parentheses 16 open parentheses x minus 1 close parentheses squared close parentheses end fraction close parenthesesequals 9 over 16 times 1 times 1 times 1equals 9 over 16 end style$

Jadi, jawaban yang tepat adalah E.