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Pertanyaan

Nilai x → 1 lim ​ 2 + 2 x ​ − 6 − 2 x ​ x 3 − x 2 ​ adalah ...

Nilai  adalah ...

  1. negative 2

  2. negative 1

  3. 0

  4. 1

  5. 2

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D. Enty

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D

jawaban yang tepat adalah D

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Pembahasan

Pembahasan
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Substitusi x = 1: Karena merupakan bentuk tak tentu, maka tidak boleh. Jadi harus diselesaiakan dengan metode mengalikan akar sekawan: (ingat: ) Jadi, jawaban yang tepat adalah D

Substitusi x = 1:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction end cell equals cell fraction numerator 1 cubed minus 1 squared over denominator square root of 2 plus 2 times 1 end root minus square root of 6 minus 2 times 1 end root end fraction end cell row blank equals cell fraction numerator 1 minus 1 over denominator square root of 4 minus square root of 4 end fraction end cell row blank equals cell 0 over 0 space open parentheses tak space tentu close parentheses end cell end table

Karena 0 over 0 merupakan bentuk tak tentu, maka tidak boleh. Jadi harus diselesaiakan dengan metode mengalikan akar sekawan: (ingat: open parentheses a plus b close parentheses open parentheses a minus b close parentheses equals a squared minus b squared)

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction end cell end table equals limit as x rightwards arrow 1 of fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction cross times fraction numerator square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root over denominator square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator open parentheses square root of 2 plus 2 x end root close parentheses squared minus open parentheses square root of 6 minus 2 x end root close parentheses squared end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator left parenthesis 2 plus 2 x right parenthesis minus left parenthesis 6 minus 2 x right parenthesis end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 2 plus 2 x minus 6 plus 2 x end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 x minus 4 end fraction equals limit as x rightwards arrow 1 of fraction numerator x squared up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 up diagonal strike left parenthesis x minus 1 right parenthesis end strike end fraction equals limit as x rightwards arrow 1 of fraction numerator x squared open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 end fraction equals fraction numerator 1 squared open parentheses square root of 2 plus 2 left parenthesis 1 right parenthesis end root plus square root of 6 minus 2 left parenthesis 1 right parenthesis end root close parentheses over denominator 4 end fraction equals fraction numerator square root of 4 plus square root of 4 over denominator 4 end fraction equals fraction numerator 2 plus 2 over denominator 4 end fraction equals 4 over 4 equals 1 

Jadi, jawaban yang tepat adalah D

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Anisyah Ramadhani

Mudah dimengerti

Eduward Jhon Bryan Gultom

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