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Pertanyaan

N i l ai → ∞ lim ​ ( 4 x 2 + 3 x + 4 ​ − 2 x + 1 ) = ...

  1. negative 7 over 4

  2. 0

  3. 3 over 4

  4. 7 over 4

  5. infinity

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F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

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Pembahasan

P e n y e l e s a i a n space l i m i t space t a k space h i n g g a space b e r b e n t u k  limit as x rightwards arrow infinity of open parentheses square root of a x squared plus b x plus c end root minus square root of p x squared plus q x plus r end root close parentheses  a d a l a h  fraction numerator b minus q over denominator 2 square root of a end fraction  M a k a semicolon  limit as x rightwards arrow infinity of open parentheses square root of 4 x squared plus 3 x plus 4 end root minus square root of 4 x squared minus 4 x plus 1 end root close parentheses    equals fraction numerator 3 minus open parentheses negative 4 close parentheses over denominator 2 square root of 4 end fraction equals 7 over 4

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Nilai dari L i m x → ∞ ​ ( 7 x − 3 ​ − 7 x + 1 ​ ) =

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