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Pertanyaan

limit as x rightwards arrow 4 of fraction numerator open parentheses x minus 4 close parentheses over denominator square root of x minus 2 end fraction equals....

Pembahasan Soal:

Limit fungsi dengan metode mengalikan akar sekawan.

Cek limit terlebih dahulu jika nilai limit 0 over 0 maka harus disederhanakan dengan pemfaktoran atau mengalikan dengan akar sekawan.

limit as x rightwards arrow 4 of space fraction numerator open parentheses x minus 4 close parentheses over denominator square root of x minus 2 end fraction equals fraction numerator open parentheses 4 minus 4 close parentheses over denominator square root of 4 minus 2 end fraction equals fraction numerator 4 minus 4 over denominator 2 minus 2 end fraction equals 0 over 0

Karena nilai limit 0 over 0, dan terdapat bentuk akar maka disederhanakan dengan cara mengalikan dengan akar sekawan.

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

Jadi, diperoleh limit as x rightwards arrow 4 of space fraction numerator open parentheses x minus 4 close parentheses over denominator square root of x minus 2 end fraction equals 4.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Dwi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 02 Mei 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.

Roboguru

Nilai dari

Pembahasan Soal:

Dengan metode perkalian sekawan, maka penyelesaiannya adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 0 of fraction numerator 3 minus square root of 2 straight x plus 9 end root over denominator straight x end fraction end cell equals cell limit as straight x rightwards arrow 0 of fraction numerator 3 minus square root of 2 straight x plus 9 end root over denominator straight x end fraction times fraction numerator left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis over denominator left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of fraction numerator 9 minus left parenthesis 2 straight x plus 9 right parenthesis over denominator straight x left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of fraction numerator 9 minus 2 straight x minus 9 over denominator straight x left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of fraction numerator negative 2 straight x over denominator straight x left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of fraction numerator negative 2 over denominator left parenthesis 3 plus square root of 2 straight x plus 9 end root right parenthesis end fraction end cell row blank equals cell fraction numerator negative 2 over denominator left parenthesis 3 plus square root of 2 times 0 plus 9 end root right parenthesis end fraction end cell row blank equals cell fraction numerator negative 2 over denominator left parenthesis 3 plus square root of 9 right parenthesis end fraction end cell row blank equals cell fraction numerator negative 2 over denominator left parenthesis 3 plus 3 right parenthesis end fraction end cell row blank equals cell fraction numerator negative 2 over denominator 6 end fraction end cell row blank equals cell negative 1 third end cell end table end style 

Jadi, Nilai begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow 0 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 minus square root of 2 straight x plus 9 end root over denominator straight x end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 third end cell end table end style 

Roboguru

Pembahasan Soal:

Nilai dari limit fungsi aljabar tersebut dapat ditentukan dengan metode perkalian akar sekawan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 2 of space fraction numerator 2 minus square root of 2 minus x end root over denominator 6 plus x minus x squared end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space fraction numerator 2 minus square root of 2 minus x end root over denominator 6 plus x minus x squared end fraction cross times fraction numerator 2 plus square root of 2 minus x end root over denominator 2 plus square root of 2 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space fraction numerator 4 minus open parentheses 2 minus x close parentheses over denominator open parentheses 3 minus x close parentheses open parentheses 2 plus x close parentheses open parentheses 2 plus square root of 2 minus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space fraction numerator 2 plus x over denominator open parentheses 3 minus x close parentheses open parentheses 2 plus x close parentheses open parentheses 2 plus square root of 2 minus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space fraction numerator 1 over denominator open parentheses 3 minus x close parentheses open parentheses 2 plus square root of 2 minus x end root close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses 3 plus 2 close parentheses open parentheses 2 plus square root of 4 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator 5 times 4 end fraction end cell row blank equals cell 1 over 20 end cell end table

Dengan demikian, nilai limit as x rightwards arrow negative 2 of space fraction numerator 2 minus square root of 2 minus x end root over denominator 6 plus x minus x squared end fraction equals 1 over 20 

Roboguru

Pembahasan Soal:

Perhitungan limit fungsi begin mathsize 14px style f open parentheses x close parentheses space text untuk  end text x text →a, a≠0 end text end style dapat dilakukan dengan metode mengalikan akar sekawan.

Dengan mengalikan akar sekawan

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

Jadi, begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator x squared minus 9 over denominator square root of x squared plus 7 end root minus 4 end fraction end style adalah 8.

Roboguru

Nilai

Pembahasan Soal:

limit as x rightwards arrow 3 of open parentheses fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction cross times fraction numerator 2 plus square root of x plus 1 end root over denominator 2 plus square root of x plus 1 end root end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator 4 minus open parentheses x plus 1 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator 4 minus x minus 1 over denominator open parentheses x minus 3 close parentheses open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator 3 minus x over denominator open parentheses x minus 3 close parentheses open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator negative open parentheses x minus 3 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction close parentheses equals limit as x rightwards arrow 3 of open parentheses fraction numerator negative 1 over denominator open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction close parentheses equals fraction numerator negative 1 over denominator 2 plus square root of 3 plus 1 end root end fraction equals fraction numerator negative 1 over denominator 2 plus 2 end fraction equals negative 1 fourth 

Dengan demikian, hasil dari limit as x rightwards arrow 3 of open parentheses fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction close parentheses equals negative 1 fourth.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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