Roboguru

Tentukan x→0lim​2−4−x​x​.

Pertanyaan

Tentukan limit as x rightwards arrow 0 of fraction numerator x over denominator 2 minus square root of 4 minus x end root end fraction.space 

Pembahasan Soal:

Dengan substitusi langsung:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator x over denominator 2 minus square root of 4 minus x end root end fraction end cell equals cell fraction numerator 0 over denominator 2 minus square root of 4 minus 0 end root end fraction end cell row blank equals cell fraction numerator 0 over denominator 2 minus 2 end fraction end cell row blank equals cell 0 over 0 space open parentheses tak space tentu close parentheses end cell end table 

Dengan mengalikan bentuk sekawan:

Bentuk 2 minus square root of 4 minus x end root mempunyai bentuk sekawan 2 plus square root of 4 minus x end root.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of fraction numerator x over denominator 2 minus square root of 4 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator x over denominator 2 minus square root of 4 minus x end root end fraction times fraction numerator 2 plus square root of 4 minus x end root over denominator 2 plus square root of 4 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator x open parentheses 2 plus square root of 4 minus x end root close parentheses over denominator 4 minus open parentheses 4 minus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator x open parentheses 2 plus square root of 4 minus x end root close parentheses over denominator x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses 2 plus square root of 4 minus x end root close parentheses end cell row blank equals cell 2 plus square root of 4 minus 0 end root end cell row blank equals cell 2 plus 2 end cell row blank equals 4 end table 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as 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 x over denominator 2 minus square root of 4 minus x end root end fraction end cell end table equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Jika  maka x→0lim​f(x)=...

Pembahasan Soal:

Ingat jika mengerjakan limit dengan substitusi langsung dan hasilnya 0 over 0 atauspace infinity over infinity maka dapat menggunakan perkalian dengan sekawan yang bertujuan untuk mengubah bentuk suatu fungsi agar ketika dilakukan substitusi dihasilkan suatu nilai. Sehingga perhitunganya sebagai berikut.

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

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

x→3lim​x​−3​(x−3)(x​+3​)​=....

Pembahasan Soal:

Cara paling dasar dalam menentukan nilai limit fungsi aljabar adalah dengan metode substitusi. Apabila hasil dari substitusi tersebut menghasilkan bentuk tak tentu, penyelesaian limit harus menggunakan metode lain, yaitu pemfaktoran atau perkalian akar sekawan.

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

Hasil dari substitusi tersebut merupakan bentuk tak tentu sehingga penyelesaian soal limit tersebut menggunakan perkalian akar sekawan sebagai berikut.

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

Dengan demikian, nilai dari table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator left parenthesis x minus 3 right parenthesis left parenthesis square root of x plus square root of 3 right parenthesis over denominator square root of x minus square root of 3 end fraction end cell equals 12 end table 

 

2

Roboguru

Hitunglah nilai setiap limit berikut. b. x→0lim​2+x​−2−x​1+x​−1−x​​

Pembahasan Soal:

Perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator square root of 1 plus x end root minus square root of 1 minus x end root over denominator square root of 2 plus x end root minus square root of 2 minus x end root end fraction end cell equals cell fraction numerator square root of 1 plus 0 end root minus square root of 1 minus 0 end root over denominator square root of 2 plus 0 end root minus square root of 2 minus 0 end root end fraction end cell row blank equals cell fraction numerator square root of 1 minus square root of 1 over denominator square root of 2 minus square root of 2 end fraction end cell row blank equals cell 0 over 0 end cell end table 

Karena hasil limit tersebut ketika disubstitusikan x equals 0 menghasil bilangan tak tentu 0 over 0, maka dilakukan rasionalisasi bentuk 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 0 of fraction numerator square root of 1 plus x end root minus square root of 1 minus x end root over denominator square root of 2 plus x end root minus square root of 2 minus x end root end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator square root of 1 plus x end root minus square root of 1 minus x end root over denominator square root of 2 plus x end root minus square root of 2 minus x end root end fraction times fraction numerator square root of 1 plus x end root plus square root of 1 minus x end root over denominator square root of 1 plus x end root plus square root of 1 minus x end root end fraction times fraction numerator square root of 2 plus x end root plus square root of 2 minus x end root over denominator square root of 2 plus x end root plus square root of 2 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses up diagonal strike 1 plus x up diagonal strike negative 1 end strike plus x close parentheses open parentheses square root of 2 plus x end root plus square root of 2 minus x end root close parentheses over denominator open parentheses up diagonal strike 2 plus x up diagonal strike negative 2 end strike plus x close parentheses open parentheses square root of 1 plus x end root plus square root of 1 minus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses up diagonal strike 2 x end strike close parentheses open parentheses square root of 2 plus x end root plus square root of 2 minus x end root close parentheses over denominator open parentheses up diagonal strike 2 x end strike close parentheses open parentheses square root of 1 plus x end root plus square root of 1 minus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses square root of 2 plus x end root plus square root of 2 minus x end root close parentheses over denominator open parentheses square root of 1 plus x end root plus square root of 1 minus x end root close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses square root of 2 plus 0 end root plus square root of 2 minus 0 end root close parentheses over denominator open parentheses square root of 1 plus 0 end root plus square root of 1 minus 0 end root close parentheses end fraction end cell row blank equals cell fraction numerator square root of 2 plus square root of 2 over denominator square root of 1 plus square root of 1 end fraction end cell row blank equals cell fraction numerator up diagonal strike 2 square root of 2 over denominator up diagonal strike 2 square root of 1 end fraction end cell row blank equals cell fraction numerator square root of 2 over denominator 1 end fraction end cell row blank equals cell square root of 2 end cell end table end style 

Jadi, limit as x rightwards arrow 0 of fraction numerator square root of 1 plus x end root minus square root of 1 minus x end root over denominator square root of 2 plus x end root minus square root of 2 minus x end root end fraction equals square root of 2.

0

Roboguru

x→2lim​4x−83x+3​−5x−1​​=....

Pembahasan Soal:

Dengan menggunakan metode substitusi didapatkan

limx24x83x+35x1===886+310109900

Karena menghasilkan nilai 00, maka dapat diselesaikan dengan metode lain, yaitu mengalikan dengan sekawannya sebagai berikut

=======limx24x83x+35x1limx24x83x+35x13x+3+5x13x+3+5x1limx2(4x8)(3x+3+5x1)3x+3(5x1)limx24(x2)(3x+3+5x1)2x+4limx24(x2)(3x+3+5x1)2(x2)limx22(3x+3+5x1)12(3+3)1121

Dengan demikian, nilai dari x2lim4x83x+35x1=121

0

Roboguru

x→3lim​4−x2+7​9−x2​=...

Pembahasan Soal:

Gunakan konsep menentukan limit suatu fungsi dengan metode mengalikan akar sekawan

Untuk menentukan nilai limit tersebut, terlebih dahulu substitusikan nilai limit terlebih dahulu, diperoleh.

limx34x2+79x2====4(3)2+79(3)249+79944000

Diperoleh hasil 00, sehingga untuk menentukan nilai limitnya dengan cara mengalikan akar sekawan.

Perhatikan perhitungan berikut.

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

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

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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