Pertanyaan

Hasil dari adalah ....

Hasil dari $limit as x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 3 x minus 9 end fraction$ adalah ....

S. Luke

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Perhatikan perhitungan berikut! Perhatikan bahwa jika maka Oleh karena itu, didapat pula atau . Dengan demikian, diperoleh hasil perhitungan sebagai berikut. Dengan demikian, hasil dari adalah . Jadi, jawaban yang tepat adalah C.

Perhatikan perhitungan berikut!

$begin mathsize 14px style limit as x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 3 x minus 9 end fraction equals limit as x rightwards arrow 3 of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 3 x minus 9 end fraction times fraction numerator 2 x minus 6 over denominator 2 x minus 6 end fraction close parentheses equals limit as x rightwards arrow 3 of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 2 x minus 6 end fraction times fraction numerator 2 x minus 6 over denominator 3 x minus 9 end fraction close parentheses end style$

Perhatikan bahwa jika x rightwards arrow 3 comma maka x minus 3 rightwards arrow 0.

Oleh karena itu, didapat pula atau .

Dengan demikian, diperoleh hasil perhitungan sebagai berikut.

$limit as x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 3 x minus 9 end fraction equals limit as x rightwards arrow 3 of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 2 x minus 6 end fraction times fraction numerator 2 x minus 6 over denominator 3 x minus 9 end fraction close parentheses equals limit as x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 2 x minus 6 end fraction times limit as x rightwards arrow 3 of invisible function application fraction numerator 2 x minus 6 over denominator 3 x minus 9 end fraction equals limit as open parentheses 2 x minus 6 close parentheses rightwards arrow 0 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 2 x minus 6 end fraction times limit as x rightwards arrow 3 of invisible function application fraction numerator 2 open parentheses x minus 3 close parentheses over denominator 3 open parentheses x minus 3 close parentheses end fraction equals limit as open parentheses 2 x minus 6 close parentheses rightwards arrow 0 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 2 x minus 6 end fraction times limit as x rightwards arrow 3 of invisible function application fraction numerator 2 up diagonal strike open parentheses x minus 3 close parentheses end strike over denominator 3 up diagonal strike open parentheses x minus 3 close parentheses end strike end fraction equals 1 times 2 over 3 equals 2 over 3$

Dengan demikian, hasil dari $limit as x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 x minus 6 close parentheses over denominator 3 x minus 9 end fraction$ adalah 2 over 3 .

Jadi, jawaban yang tepat adalah C.