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Pertanyaan

Hasil dari x → 4 lim ​ x − 4 2 x + 1 ​ − x + 5 ​ ​ = ...

Hasil dari 

  1. 1 over 6

  2. 0

  3. 6

  4. 2

  5. 1 half

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G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

jawaban yang benar adalah A.

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Pembahasan

Untuk mencari nilai limit, maka dengan menggunakan perkalian akar sekawan, sehingga berlaku aturan, Dengan dan . Maka, Maka di dapat nilai limit dari adalah . Oleh karena itu, jawaban yang benar adalah A.

Untuk mencari nilai limit, maka dengan menggunakan perkalian akar sekawan, sehingga berlaku aturan,

a squared minus b squared equals open parentheses a plus b close parentheses open parentheses a minus b close parentheses

Dengan a equals square root of 2 x plus 1 end root dan b equals square root of x plus 5 end root. Maka,

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator square root of 2 x plus 1 end root minus square root of x plus 5 end root over denominator x minus 4 end fraction cross times fraction numerator square root of 2 x plus 1 end root plus square root of x plus 5 end root over denominator square root of 2 x plus 1 end root plus square root of x plus 5 end root end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator 2 x plus 1 minus open parentheses x plus 5 close parentheses over denominator open parentheses x minus 4 close parentheses open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator 2 x plus 1 minus x minus 5 over denominator open parentheses x minus 4 close parentheses open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator 2 x minus x plus 1 minus 5 over denominator open parentheses x minus 4 close parentheses open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator up diagonal strike x minus 4 end strike over denominator up diagonal strike open parentheses x minus 4 close parentheses end strike open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 1 over denominator open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses square root of 2 left parenthesis 4 right parenthesis plus 1 end root plus square root of 4 plus 5 end root close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 8 plus 1 end root plus square root of 9 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 9 plus square root of 9 end fraction end cell row blank equals cell fraction numerator 1 over denominator 3 plus 3 end fraction end cell row blank equals cell 1 over 6 end cell end table end style

Maka di dapat nilai limit dari limit as x rightwards arrow 4 of fraction numerator square root of 2 x plus 1 end root minus square root of x plus 5 end root over denominator x minus 4 end fraction adalah 1 over 6.


Oleh karena itu, jawaban yang benar adalah A.

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