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Pertanyaan

Nilai x → 1 lim ​ x 2 − 1 5 x − 4 ​ − x ​ = ....

Nilai 

  1. 1 fourth space

  2. 1 half space

  3. 3 over 4 space

  4. 2 space

  5. 4 space

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R. Febrianti

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

jawaban yang benar adalah C.

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Pembahasan

Nilai dapat ditentukan dengan mensubstitusi nilai ke dalam fungsi sehingga Karena hasil menggunakan konsep limit yaitu cara substitusi adalah maka untuk menentukan limitnya menggunakan cara selanjutnya yaitumengalikan dengan akar sekawansebagai berikut Oleh karena itu, jawaban yang benar adalah C.

Nilai limit as x rightwards arrow 1 of fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction dapat ditentukan dengan mensubstitusi nilai x equals 1 ke dalam fungsi fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction end cell equals cell fraction numerator square root of 5 open parentheses 1 close parentheses minus 4 end root minus open parentheses 1 close parentheses over denominator open parentheses 1 close parentheses squared minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction end cell equals cell fraction numerator square root of 1 minus 1 over denominator 1 minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction end cell equals cell 0 over 0 end cell end table

Karena hasil limit as x rightwards arrow 1 of fraction numerator square root of 5 x minus 4 end root minus x over denominator x squared minus 1 end fraction menggunakan konsep limit yaitu cara substitusi adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 0 over 0 comma end cell end table maka untuk menentukan limitnya menggunakan cara selanjutnya yaitu mengalikan dengan akar sekawan sebagai berikut

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


Oleh karena itu, jawaban yang benar adalah C.

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Nasywa Qothrunnada Kamilah

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