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Nilai x → 1 lim ​ 1 − x ( 5 − x ​ − 2 ) ( 2 − x ​ + 1 ) ​ adalah ....

Nilai  adalah ....

  1. begin mathsize 14px style negative 1 half end style

  2. begin mathsize 14px style negative 1 fourth end style

  3. begin mathsize 14px style 1 over 8 end style

  4. begin mathsize 14px style 1 fourth end style

  5. begin mathsize 14px style 1 half end style

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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Pembahasan

begin mathsize 14px style limit as straight x rightwards arrow 1 of invisible function application space fraction numerator open parentheses square root of 5 minus straight x end root minus 2 close parentheses open parentheses square root of 2 minus straight x end root plus 1 close parentheses over denominator 1 minus straight x end fraction  limit as straight x rightwards arrow 1 of invisible function application space fraction numerator open parentheses square root of 5 minus straight x end root minus 2 close parentheses over denominator 1 minus straight x end fraction. space limit as straight x rightwards arrow 1 of open parentheses square root of 2 minus straight x end root plus 1 close parentheses  limit as straight x rightwards arrow 1 of invisible function application space fraction numerator open parentheses square root of 5 minus straight x end root minus 2 close parentheses over denominator 1 minus straight x end fraction. open parentheses square root of 2 minus 1 end root plus 1 close parentheses  2. space limit as straight x rightwards arrow 1 of invisible function application space fraction numerator open parentheses square root of 5 minus straight x end root minus 2 close parentheses over denominator 1 minus straight x end fraction    Gunakan space dalil space straight L space hospital  2. space limit as straight x rightwards arrow 1 of invisible function application space fraction numerator negative fraction numerator 1 over denominator 2 square root of 5 minus straight x end root end fraction over denominator negative 1 end fraction  2. fraction numerator 1 over denominator 2 square root of 4 end fraction equals 1 half end style

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Jika f ( x ) = x 1 ​ maka nilai h → 0 lim ​ h f ( x + h ) f ( h ) ​ adalah ...

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