Roboguru

Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction equals... end style

Pembahasan Soal:

dengan substitusi nilai x, diperoleh bentuk tak tentu 0/0, sehingga dengan mengalikan sekawan

begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction cross times fraction numerator 4 plus square root of x squared plus 7 end root over denominator 4 plus square root of x squared plus 7 end root end fraction equals limit as x rightwards arrow 3 of fraction numerator open parentheses 9 minus x squared close parentheses open parentheses 4 plus square root of x squared plus 7 end root close parentheses over denominator 16 minus open parentheses x squared plus 7 close parentheses end fraction equals limit as x rightwards arrow 3 of fraction numerator left parenthesis 9 minus x squared right parenthesis open parentheses 4 plus square root of x squared plus 7 end root close parentheses over denominator 9 minus x squared end fraction equals limit as x rightwards arrow 3 of left parenthesis 4 plus square root of x squared plus 7 end root right parenthesis equals left parenthesis 4 plus square root of 3 squared plus 7 end root right parenthesis equals 8 end style

Jadi, hasilnya adalah 8.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Janatu

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 30 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai  adalah ....

Pembahasan Soal:

Dengan mengalikan akar sekawan.

begin mathsize 14px style limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction times fraction numerator square root of 12 plus square root of x plus 8 end root over denominator square root of 12 plus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator 12 minus open parentheses x plus 8 close parentheses end fraction equals limit as x rightwards arrow 4 of fraction numerator up diagonal strike open parentheses 4 minus x close parentheses end strike open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator up diagonal strike open parentheses 4 minus x close parentheses end strike end fraction equals limit as x rightwards arrow 4 of square root of 12 plus square root of x plus 8 end root equals square root of 12 plus square root of 4 plus 8 end root equals square root of 12 plus square root of 12 equals 2 square root of 12 equals 2 square root of 4 times 3 end root equals 2 times 2 square root of 3 equals 4 square root of 3 end style 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Nilai

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction end cell equals cell fraction numerator 9 minus open parentheses 3 close parentheses squared over denominator 4 minus square root of open parentheses 3 close parentheses squared plus 7 end root end fraction end cell row blank equals cell fraction numerator 9 minus 9 over denominator 4 minus square root of 16 end fraction end cell row blank equals cell 0 over 0 end cell end table
 

Berdasarkan uraian di atas, dengan metode substitusi langsung, diperoleh hasil limitnya adalah 0 over 0, maka nilai limit dapat ditentukan dengan metode mengalikan akar sekawan terlebih dahulu sebagai berikut.
 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction end cell equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction cross times fraction numerator 4 plus square root of x squared plus 7 end root over denominator 4 plus square root of x squared plus 7 end root end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator 9 minus x squared left parenthesis 4 plus square root of x squared plus 7 end root right parenthesis over denominator 16 minus left parenthesis x squared plus 7 right parenthesis end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator up diagonal strike 9 minus x squared end strike left parenthesis 4 plus square root of x squared plus 7 end root right parenthesis over denominator up diagonal strike 9 minus x squared end strike end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses 4 plus square root of x squared plus 7 end root close parentheses end cell row blank equals cell 4 plus square root of open parentheses 3 close parentheses squared plus 7 end root end cell row blank equals cell 4 plus 4 end cell row blank equals 8 end table


Jadi, limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction equals 8.

0

Roboguru

Nilai  adalah ....

Pembahasan Soal:

Kita coba dengan mensubstitusikan nilai begin mathsize 14px style x equals 7 end style ke begin mathsize 14px style limit as x rightwards arrow 7 of open parentheses fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction close parentheses end style, sehingga

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 7 of open parentheses fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction close parentheses end cell equals cell fraction numerator 2 open parentheses 7 close parentheses squared minus 15 left parenthesis 7 right parenthesis plus 7 over denominator square root of 7 plus 2 end root minus 3 end fraction end cell row blank equals cell fraction numerator 98 minus 105 plus 7 over denominator 3 minus 3 end fraction end cell row blank equals cell 0 over 0 end cell end table end style 

Berdasarkan perhitungan, didapat hasil undefined adalah dalam bentuk tak tentu.

Untuk menyelesaikan soal tersebut, kita bisa coba dengan memanipulasi soal/pernyataan yang diberikan.

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 7 of open parentheses fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction cross times fraction numerator square root of x plus 2 end root plus 3 over denominator square root of x plus 2 end root plus 3 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator open parentheses 2 x squared minus 15 x plus 7 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator open parentheses square root of x plus 2 end root close parentheses squared minus 3 squared end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator open parentheses 2 x squared minus 15 x plus 7 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator x plus 2 minus 9 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x minus 7 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator x minus 7 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses up diagonal strike x minus 7 end strike close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator up diagonal strike x minus 7 end strike end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 7 of open parentheses open parentheses 2 x minus 1 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses close parentheses end cell end table end style 

Setelah kita manipulasi dan dapat bentuk paling sederhana, kita substitusikan nilai undefined.

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 7 of open parentheses fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction close parentheses end cell equals cell limit as x rightwards arrow 7 of open parentheses open parentheses 2 x minus 1 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses close parentheses end cell row blank equals cell open parentheses open parentheses 2 open parentheses 7 close parentheses minus 1 close parentheses open parentheses square root of 7 plus 2 end root plus 3 close parentheses close parentheses end cell row blank equals cell open parentheses 14 minus 1 close parentheses open parentheses square root of 9 plus 3 close parentheses end cell row blank equals cell open parentheses 13 close parentheses open parentheses 3 plus 3 close parentheses end cell row blank equals cell open parentheses 13 close parentheses open parentheses 6 close parentheses end cell row blank equals 78 end table end style 

Jadi, nilai dari begin mathsize 14px style stack bold l bold i bold m with bold x bold rightwards arrow bold 7 below bold space fraction numerator bold 2 bold x to the power of bold 2 bold minus bold 15 bold x bold plus bold 7 over denominator square root of bold x bold plus bold 2 end root bold minus bold 3 end fraction end style adalah begin mathsize 14px style bold 78 end style.undefined

0

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of fraction numerator square root of 6 x minus 2 end root minus square root of 3 x plus 7 end root over denominator x minus 3 end fraction space end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator square root of 6 x minus 2 end root minus square root of 3 x plus 7 end root over denominator x minus 3 end fraction cross times fraction numerator square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root over denominator square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root end fraction close parentheses space space end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator left parenthesis 6 x minus 2 right parenthesis minus left parenthesis 3 x plus 7 right parenthesis over denominator left parenthesis x minus 3 right parenthesis left parenthesis square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root right parenthesis end fraction space space end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 3 x minus 9 over denominator left parenthesis x minus 3 right parenthesis left parenthesis square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root right parenthesis end fraction space space end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 3 left parenthesis x minus 3 right parenthesis over denominator left parenthesis x minus 3 right parenthesis left parenthesis square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root right parenthesis end fraction space space end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 3 over denominator left parenthesis square root of 6 x minus 2 end root plus square root of 3 x plus 7 end root right parenthesis end fraction space space end cell row blank equals cell fraction numerator 3 over denominator square root of left parenthesis 6 cross times 3 right parenthesis minus 2 end root plus square root of left parenthesis 3 cross times 3 right parenthesis plus 7 end root end fraction space end cell row space equals cell fraction numerator 3 over denominator square root of 18 minus 2 end root plus square root of 9 plus 7 end root end fraction space end cell row space equals cell fraction numerator 3 over denominator square root of 16 plus square root of 16 end fraction space end cell row space equals cell fraction numerator 3 over denominator 4 plus 4 end fraction space space end cell row blank equals cell 3 over 8 end cell end table end style  

Jadi, jawaban yang tepat adalah D.undefined 

0

Roboguru

Nilai  adalah ....

Pembahasan Soal:

Dengan menerapkan penyelesaian limit dengan metode mengalikan akar sekawan dan metode pemfaktoran, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 7 of space fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 7 of space fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction cross times fraction numerator square root of x plus 2 end root plus 3 over denominator square root of x plus 2 end root plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 7 of space fraction numerator open parentheses 2 x squared minus 15 x plus 7 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator open parentheses square root of x plus 2 end root close parentheses squared minus 3 squared end fraction end cell row blank equals cell limit as x rightwards arrow 7 of space fraction numerator open parentheses 2 x squared minus 15 x plus 7 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator x plus 2 minus 9 end fraction end cell row blank equals cell limit as x rightwards arrow 7 of space fraction numerator open parentheses 2 x minus 1 close parentheses up diagonal strike open parentheses x minus 7 close parentheses end strike open parentheses square root of x plus 2 end root plus 3 close parentheses over denominator up diagonal strike x minus 7 end strike end fraction end cell row blank equals cell limit as x rightwards arrow 7 of space open parentheses 2 x minus 1 close parentheses open parentheses square root of x plus 2 end root plus 3 close parentheses end cell row blank equals cell open parentheses 2 times 7 minus 1 close parentheses open parentheses square root of 7 plus 2 end root plus 3 close parentheses end cell row blank equals cell 13 times 6 end cell row blank equals 78 end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 7 of space fraction numerator 2 x squared minus 15 x plus 7 over denominator square root of x plus 2 end root minus 3 end fraction end style adalah 78.

Dengan demikian, jawaban yang tepat adalah A.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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