Pertanyaan

Nilai dari adalah ....

Nilai dari $begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator square root of 2 minus square root of 2 plus 3 over x end root end fraction end style$ adalah ....

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H. Nufus

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Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

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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 tan invisible function application open parentheses 6 over x close parentheses over denominator square root of 2 minus square root of 2 plus 3 over x end root end fraction equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator square root of 2 minus square root of 2 plus 3 over x end root end fraction times fraction numerator square root of 2 plus square root of 2 plus 3 over x end root over denominator square root of 2 plus square root of 2 plus 3 over x end root end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator 2 minus open parentheses 2 plus 3 over x close parentheses end fraction times open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times limit as x rightwards arrow straight infinity of invisible function application open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses square root of 2 plus square root of 2 plus 0 end root close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses 2 square root of 2 close parentheses end style$

Misalkan . Ketika , maka . Sehingga

Jadi, jawaban yang tepat adalah E.