Roboguru

Tentukan nilai dari:

Pertanyaan

Tentukan nilai dari: 

begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end style   

Pembahasan Soal:

Tentukan nilai limit dengan mensubstitusikan nilai x=3 ke dalam fungsi.

limx3x32x+1===3323+1332400

Karena dengan cara substitusi menghasilkan nilai 00, maka perlu kita gunakan cara lain yakni dengan mengalikan dengan 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 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end cell equals cell 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 times open parentheses 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 end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 2 minus square root of x plus 1 end root close parentheses open parentheses 2 plus square root of x plus 1 end root 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 end cell row blank equals cell limit as x rightwards arrow 3 of 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 end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 3 minus x 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 end cell row blank equals cell limit as x rightwards arrow 3 of 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 end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator negative 1 over denominator open parentheses 2 plus square root of x plus 1 end root close parentheses end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 2 plus square root of 3 plus 1 end root end fraction end cell row blank equals cell negative fraction numerator 1 over denominator 2 plus 2 end fraction end cell row blank equals cell negative 1 fourth end cell end table end style 

Dengan demikian, nilai begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction equals negative 1 fourth end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Find the value of the limit.

Pembahasan Soal:

Dengan mengalikan akar sekawan, diperoleh:

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

Sehingga,

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

0

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Diketahui begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end style. Untuk x equals 3, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end cell equals cell fraction numerator 2 minus square root of 3 plus 1 end root over denominator 3 minus 3 end fraction end cell row blank equals cell fraction numerator 2 minus square root of 4 over denominator 3 minus 3 end fraction end cell row blank equals cell fraction numerator 2 minus 2 over denominator 3 minus 3 end fraction end cell row blank equals cell 0 over 0 end cell end table

Karena untuk x equals 3, diperoleh nilai limit sebesar 0 over 0, maka untuk penyelesaian limit tersebut digunakan perkalian sekawan. Diperoleh

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

Dengan demikian, nilai dari begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end style adalah negative 1 fourth.

0

Roboguru

Hitunglah nilai setiap limit berikut. b.

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

Pembahasan Soal:

Cara utama dalam menentukan nilai limit adalah dengan menggunakan metode substitusi. Apabila hasil substitusi tersebut merupakan bentuk tak tentu, selanjutnya penyelesaian harus menggunakan cara lain, yaitu metode pemfaktoran atau kali akar sekawan.

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

Hasil dari substitusi merupakan bentuk tak tentu sehingga penyelesaian limit tersebut menggunakan perkalian akar sekawan 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 4 of fraction numerator square root of 2 x plus 1 end root minus square root of x plus 5 end root over denominator 4 minus x end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator square root of 2 x plus 1 end root minus square root of x plus 5 end root over denominator 4 minus x end fraction cross times fraction numerator square root of 2 x plus 1 end root plus square root of x plus 5 end root over denominator square root of 2 x plus 1 end root plus square root of x plus 5 end root end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator 2 x plus 1 minus left parenthesis x plus 5 right parenthesis over denominator open parentheses 4 minus x close parentheses open parentheses square root of 2 x plus 1 end root plus square root of x plus 5 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator open parentheses 4 minus x close parentheses left parenthesis square root of 2 x plus 1 end root plus square root of x plus 5 end root right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator negative 1 over denominator left parenthesis square root of 2 x plus 1 end root plus square root of x plus 5 end root right parenthesis end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 3 plus 3 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 6 end fraction end cell end table end style 

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator square root of 2 x plus 1 end root minus square root of x plus 5 end root over denominator 4 minus x end fraction end cell equals cell negative 1 over 6 end cell end table 

0

Roboguru

Tentukan nilai dari:

Pembahasan Soal:

Cara perkalian akar sekawan dipakai jika hasil uji substitusi menghasilkan bentuk tak tentu, dan khusus untuk soal limit yang fungsinya berbentuk akar.

Pertama, kita uji terlebih dahulu dengan menggunakan cara substitusi.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 16 of fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell equals cell fraction numerator square root of 16 minus 4 over denominator 16 minus 16 end fraction end cell row blank equals cell 0 over 0 end cell end table 

Karena menghasilkan bentuk tak tentu, maka kita gunakan cara perkalian akar sekawan seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 16 of fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell equals cell limit as x rightwards arrow 16 of open parentheses fraction numerator square root of x minus 4 over denominator x minus 16 end fraction close parentheses times open parentheses fraction numerator square root of x plus 4 over denominator square root of x plus 4 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 16 of fraction numerator open parentheses x minus 16 close parentheses over denominator open parentheses x minus 16 close parentheses open parentheses square root of x plus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 16 of fraction numerator 1 over denominator square root of x plus 4 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 16 plus 4 end fraction end cell row blank equals cell fraction numerator 1 over denominator 4 plus 4 end fraction end cell row blank equals cell 1 over 8 end cell end table end style  

Jadi begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 16 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell end table equals 1 over 8 end style

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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