Roboguru

Find the value of the limit.

Pertanyaan

Find the value of the limit.

begin mathsize 14px style 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 style 

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

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Putri

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 14 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan nilai dari:

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

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

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

....

Pembahasan Soal:

Coba substitusikan nilai ke dalam limitnya sehingga diperoleh

x1limx1x31=11(1)31=00

Karena bernilai 00 maka harus dilakukan manipulasi pada limit tersebut dengan metode pemfaktoran. Maka

limx1x1x31====limx1x1(x2+x+1)(x1)limx1x2+x+1(1)2+(1)+13

Dengan demikian, nilai dari limx1x1x31=3.

0

Roboguru

Hitunglah nilai a dan b yang memenuhi kesamaan limit berikut. a.

Pembahasan Soal:

Jika nilai x equals 1 disubstitusikan ke dalam fungsi, maka akan menghasilkan nilai 0, maka  pembilang dan penyebut harus difaktorkan. Untuk itu, dilakukan bentuk 0 over 0 sebagai bentuk tak tentu suatu limit. limit as x rightwards arrow 1 of fraction numerator x squared plus a x plus b over denominator x minus 1 end fraction equals 0 over 0, hal ini berarti pembilang sama dengan 0 karena penyebut bernilai nol, diperoleh:

straight x equals 1 space space rightwards double arrow open parentheses 1 close parentheses squared plus straight a open parentheses 1 close parentheses plus straight b equals 0 space space space space space space space space space space space space space space space space space space space space space space space space 1 plus straight a plus straight b equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight b equals negative straight a minus 1 space space space space space space space space space

Proses pemfaktoran:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 1 of fraction numerator straight x squared plus ax plus straight b over denominator straight x minus 1 end fraction end cell equals 3 row cell limit as straight x rightwards arrow 1 of fraction numerator straight x squared plus ax plus open parentheses negative straight a minus 1 close parentheses over denominator straight x minus 1 end fraction end cell equals 3 row cell limit as straight x rightwards arrow 1 of fraction numerator straight x squared plus ax minus straight a minus 1 over denominator straight x minus 1 end fraction end cell equals 3 row cell limit as straight x rightwards arrow 1 of fraction numerator open parentheses straight x minus 1 close parentheses open parentheses straight x plus straight a plus 1 close parentheses over denominator open parentheses straight x minus 1 close parentheses end fraction end cell equals 3 row cell limit as straight x rightwards arrow 1 of fraction numerator up diagonal strike open parentheses straight x minus 1 close parentheses end strike open parentheses straight x plus straight a plus 1 close parentheses over denominator up diagonal strike open parentheses straight x minus 1 close parentheses end strike end fraction end cell equals 3 row cell straight x plus straight a plus 1 end cell equals 3 row cell open parentheses 1 close parentheses plus straight a plus 1 end cell equals 3 row cell straight a plus 2 end cell equals 3 row straight a equals 1 end table

Substitusikan straight a equals 1 ke dalam persamaan straight b equals negative straight a minus 1.

table attributes columnalign right center left columnspacing 0px end attributes row straight b equals cell negative straight a minus 1 end cell row blank equals cell negative open parentheses 1 close parentheses minus 1 end cell row blank equals cell negative 2 end cell end table

Dengan demikan nilai straight a equals 1 dan straight b equals negative 2.

0

Roboguru

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