Pertanyaan

Nilai dari adalah ....

Nilai dari $begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator square root of 5 minus 1 over x end root minus square root of 5 end fraction end style$ adalah ....

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan bahwa Misalkan . Ketika , maka . Sehingga Jadi, jawaban yang tepat adalah E.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator square root of 5 minus 1 over x end root minus square root of 5 end fraction equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator square root of 5 minus 1 over x end root minus square root of 5 end fraction times fraction numerator square root of 5 minus 1 over x end root plus square root of 5 over denominator square root of 5 minus 1 over x end root plus square root of 5 end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator open parentheses 5 minus 1 over x close parentheses minus 5 end fraction times open parentheses square root of 5 minus 1 over x end root plus square root of 5 close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator negative 1 over x end fraction times open parentheses square root of 5 minus 1 over x end root plus square root of 5 close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator negative 1 over x end fraction times limit as x rightwards arrow straight infinity of invisible function application open parentheses square root of 5 minus 1 over x end root plus square root of 5 close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator negative 1 over x end fraction times open parentheses square root of 5 minus 0 end root plus square root of 5 close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator negative 1 over x end fraction times open parentheses 2 square root of 5 close parentheses end style$

Misalkan . Ketika , maka . Sehingga

$begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application open parentheses 5 over x close parentheses over denominator negative 1 over x end fraction times open parentheses 2 square root of 5 close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator sin invisible function application 5 open parentheses 1 over x close parentheses over denominator negative 1 over x end fraction times open parentheses 2 square root of 5 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 5 y over denominator negative y end fraction times open parentheses 2 square root of 5 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 5 y over denominator negative y end fraction times 5 over 5 times open parentheses 2 square root of 5 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 5 y over denominator 5 y end fraction times fraction numerator 5 over denominator negative 1 end fraction times open parentheses 2 square root of 5 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 5 y over denominator 5 y end fraction times open parentheses negative 10 square root of 5 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 5 y over denominator 5 y end fraction times open parentheses negative 10 square root of 5 close parentheses equals 1 times open parentheses negative 10 square root of 5 close parentheses equals negative 10 square root of 5 end style$

Jadi, jawaban yang tepat adalah E.