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Nilai

Pertanyaan

Nilai limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction equals...

Pembahasan Soal:

Nilai limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction dapat ditentukan dengan mensubstitusi nilai x equals 1 ke dalam fungsi fraction numerator x squared minus 5 x plus 4 over denominator x cubed 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 x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell fraction numerator open parentheses 1 close parentheses squared minus 5 open parentheses 1 close parentheses plus 4 over denominator open parentheses 1 close parentheses cubed minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell fraction numerator 1 minus 5 plus 4 over denominator 1 minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed 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 x squared minus 5 x plus 4 over denominator x cubed 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 pemfaktoran sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed 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 open parentheses x minus 4 close parentheses over denominator up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator x minus 4 over denominator x squared plus x plus 1 end fraction end cell row blank equals cell fraction numerator 1 minus 4 over denominator open parentheses 1 close parentheses squared plus 1 plus 1 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator 3 end fraction end cell row blank equals cell negative 1 end cell end table


Jadi nilai limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction equals negative 1.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Febrianti

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 06 Juni 2021

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Pembahasan Soal:

Nilai limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses spacedapat ditentukan dengan mensubstitusi nilai x equals 1 ke dalam fungsi fraction numerator x minus 2 over denominator x plus 3 end fraction sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell fraction numerator open parentheses 1 close parentheses minus 2 over denominator open parentheses 1 close parentheses plus 3 end fraction end cell row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell fraction numerator negative 1 over denominator 4 end fraction end cell row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell negative 1 fourth end cell end table


Jadi nilai limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses spaceadalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative 1 fourth. end cell end table

Roboguru

Nilai

Pembahasan Soal:

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.

Roboguru

adalah ....

Pembahasan Soal:

Kita gunakan metode pemfaktoran,

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x squared plus 4 x minus 5 over denominator x minus 1 end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses over denominator x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of x plus 5 end cell row blank equals cell 1 plus 5 end cell row blank equals 6 end table

Jadi, begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator x squared plus 4 x minus 5 over denominator x minus 1 end fraction end style adalah 6. 

Roboguru

Pembahasan Soal:

Dengan cara pemfaktoran, maka didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 3 of fraction numerator x squared minus 6 x minus 27 over denominator x squared minus 9 end fraction end cell equals cell limit as x rightwards arrow negative 3 of fraction numerator x squared minus 6 x minus 27 over denominator x squared minus 9 end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of fraction numerator open parentheses x minus 9 close parentheses open parentheses x plus 3 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of fraction numerator open parentheses x minus 9 close parentheses over denominator open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell fraction numerator negative 3 minus 9 over denominator negative 3 minus 3 end fraction end cell row blank equals cell fraction numerator negative 12 over denominator negative 6 end fraction end cell row blank equals 2 end table

Jadi, jawaban yang tepat adalah E.

Roboguru

Pembahasan Soal:

Diketahui limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction, nilai dari limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction dapat ditentukan dengan substitusi nilai x equals 2 ke dalam fungsi fraction numerator x minus 2 over denominator x squared minus 4 end fraction sehingga diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction end cell equals cell fraction numerator 2 minus 2 over denominator 2 squared minus 4 end fraction end cell row blank equals cell fraction numerator 0 over denominator 4 minus 4 end fraction end cell row blank equals cell 0 over 0 end cell end table

Karena hasil dari limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction equals 0 over 0, maka cara menentukan nilai dari limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction dengan cara pemfaktoran seperti berikut.

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

Jadi nilai dari limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator x squared minus 4 end fraction adalah 1 over 6.

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