Pertanyaan

x → ∞ lim ( 9 x 2 + 4 x − 3 − 9 x 2 − 7 x + 2 ) = ...

$x \to \infty lim (9 x^{2} + 4 x - 3 - 9 x^{2} - 7 x + 2) = ...$

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N. Puspita

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

Pembahasan

Mencari nilai limit tak hingga: Jadi, hasilnya adalah . Dengan demikian, jawaban yang tepat adalah A.

Mencari nilai limit tak hingga:

$limit as x rightwards arrow infinity of open parentheses square root of 9 x squared plus 4 x minus 3 end root minus square root of 9 x squared minus 7 x plus 2 end root close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 9 x squared plus 4 x minus 3 end root minus square root of 9 x squared minus 7 x plus 2 end root cross times fraction numerator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator open parentheses square root of 9 x squared plus 4 x minus 3 end root close parentheses squared minus open parentheses square root of 9 x squared minus 7 x plus 2 end root close parentheses squared over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 9 x squared plus 4 x minus 3 minus open parentheses 9 x squared minus 7 x plus 2 close parentheses over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 9 x squared plus 4 x minus 3 minus 9 x squared plus 7 x minus 2 over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 9 x squared plus 4 x minus 3 minus 9 x squared plus 7 x minus 2 over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 11 x minus 5 over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 11 x minus 5 over denominator square root of 9 x squared plus 4 x minus 3 end root plus square root of 9 x squared minus 7 x plus 2 end root end fraction. fraction numerator begin display style 1 over x end style over denominator begin display style 1 over x end style end fraction close parentheses equals limit as x rightwards arrow infinity of open parentheses fraction numerator 11 minus begin display style 5 over x end style over denominator square root of 9 plus begin display style 4 over x end style minus begin display style 3 over x squared end style end root plus square root of 9 minus begin display style 7 over x end style plus begin display style 2 over x squared end style end root end fraction close parentheses equals fraction numerator 11 minus begin display style 0 end style over denominator square root of 9 plus begin display style 0 end style minus 0 end root plus square root of 9 minus begin display style 0 end style plus 0 end root end fraction equals fraction numerator 11 over denominator square root of 9 plus square root of 9 end fraction equals fraction numerator 11 over denominator 3 plus 3 end fraction equals 11 over 6$

Jadi, hasilnya adalah 11 over 6 .

Dengan demikian, jawaban yang tepat adalah A.