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Pertanyaan

Nilai x → 2 lim ​ 4 − x 2 x − − 3 x + 10 ​ ​ adalah ...

Nilai adalah ...

  1. negative 7 over 8

  2. negative 5 over 8

  3. negative 3 over 8

  4. negative 5 over 8

  5. 7 over 8

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H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

space limit as x rightwards arrow 2 of open parentheses fraction numerator x minus space square root of negative 3 x plus 10 end root over denominator 4 minus x squared end fraction close parentheses space. open parentheses fraction numerator x plus square root of negative 3 x plus 10 end root over denominator x plus square root of negative 3 x plus 10 end root end fraction close parentheses  equals space limit as x rightwards arrow 2 of space fraction numerator x squared minus open parentheses negative 3 x plus 10 close parentheses over denominator open parentheses 4 minus x squared close parentheses open parentheses x plus square root of negative 3 x plus 10 end root close parentheses end fraction  equals space limit as x rightwards arrow 2 of space fraction numerator x squared plus 3 x minus 10 over denominator negative open parentheses x squared minus 4 close parentheses open parentheses x plus square root of 3 x plus 10 end root close parentheses end fraction  equals space limit as x rightwards arrow 2 of space fraction numerator open parentheses x plus 5 close parentheses up diagonal strike open parentheses x minus 2 close parentheses end strike over denominator up diagonal strike negative open parentheses x minus 2 close parentheses end strike open parentheses x plus 2 close parentheses open parentheses x plus square root of negative 3 x plus 10 end root close parentheses end fraction  equals fraction numerator 2 plus 5 over denominator negative open parentheses 2 plus 2 close parentheses open parentheses 2 plus square root of negative 3 open parentheses 2 close parentheses plus 10 end root close parentheses end fraction  equals fraction numerator 7 over denominator 4 open parentheses 2 plus square root of 4 close parentheses end fraction equals fraction numerator 7 over denominator negative 4 open parentheses 4 close parentheses end fraction equals negative 7 over 8

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Nilai ​ ​ x → 3 l im ​ ​ x − 3 x 2 − 4 x + 3 ​ = ...

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