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Pertanyaan

Nilai dari x → ∞ lim ​ ( x 2 + x − 6 ​ − x 2 − 4 x − 12 ​ ) = ...

Nilai dari  

  1. negative 3 over 2 

  2. 2 

  3. 5 over 2 

  4. 3 

  5. 5 

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N. Puspita

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

Pembahasan

Pembahasan
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Nilai limit tersebut dapat ditentukan dengan metode perkalian akar sekawanseperti di bawa ini Dengan demikian, jawaban yang tepat adalah C.

Nilai limit tersebut dapat ditentukan dengan metode perkalian akar sekawan seperti di bawa ini 

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of open parentheses square root of x squared plus x minus 6 end root minus square root of x squared minus 4 x minus 12 end root close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of open parentheses square root of x squared plus x minus 6 end root minus square root of x squared minus 4 x minus 12 end root close parentheses cross times end cell row blank blank cell fraction numerator open parentheses square root of x squared plus x minus 6 end root plus square root of x squared minus 4 x minus 12 end root close parentheses over denominator open parentheses square root of x squared plus x minus 6 end root plus square root of x squared minus 4 x minus 12 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator open parentheses x squared plus x minus 6 close parentheses minus open parentheses x squared minus 4 x minus 12 close parentheses over denominator square root of x squared plus x minus 6 end root plus square root of x squared minus 4 x minus 12 end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 5 x plus 12 over denominator square root of x squared plus x minus 6 end root plus square root of x squared minus 4 x minus 12 end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 5 x plus 12 over denominator square root of x squared plus x minus 6 end root plus square root of x squared minus 4 x minus 12 end root end fraction cross times fraction numerator open parentheses begin display style 1 over x end style close parentheses over denominator open parentheses begin display style 1 over x end style close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 5 plus begin display style 12 over x end style over denominator square root of begin display style fraction numerator x squared plus x minus 6 over denominator x squared end fraction end style end root plus square root of begin display style fraction numerator x squared minus 4 x minus 12 over denominator x squared end fraction end style end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 5 plus begin display style 12 over x end style over denominator square root of begin display style 1 plus 1 over x minus 6 over x squared end style end root plus square root of begin display style 1 minus 4 over x minus 12 over x squared end style end root end fraction end cell row blank equals cell fraction numerator 5 plus 0 over denominator square root of 1 plus 0 minus 0 end root plus square root of 1 minus 0 minus 0 end root end fraction end cell row blank equals cell fraction numerator 5 over denominator 1 plus 1 end fraction end cell row blank equals cell 5 over 2 end cell end table 

Dengan demikian, jawaban yang tepat adalah C.

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