Roboguru

Nilai

Pertanyaan

Nilai limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction equals....

Pembahasan Soal:

Menurut konsep limit, untuk menentukan nilai suatu limit fungsi aljabar adalah dengan mensubstitusi nilai x kedalam fungsi aljabar tersebut sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell fraction numerator 8 open parentheses negative 2 close parentheses squared plus 14 open parentheses negative 2 close parentheses minus 4 over denominator 2 open parentheses negative 2 close parentheses plus 4 end fraction end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell fraction numerator 8 cross times 4 minus 28 minus 4 over denominator negative 4 plus 4 end fraction end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell 0 over 0 end cell end table


Karena hasil dari limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction ketika menggunakan cara substitusi hasilnya table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 0 over 0 comma end cell end table maka untuk menentukan hasil limitnya menggunakan cara pemfaktoran yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell limit as x rightwards arrow negative 2 of fraction numerator open parentheses 8 x minus 2 close parentheses up diagonal strike open parentheses x plus 2 close parentheses end strike over denominator 2 up diagonal strike left parenthesis x plus 2 right parenthesis end strike end fraction end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell limit as x rightwards arrow negative 2 of fraction numerator up diagonal strike 2 open parentheses 4 x minus 1 close parentheses over denominator up diagonal strike 2 end fraction end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell limit as x rightwards arrow negative 2 of open parentheses 4 x minus 1 close parentheses end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell 4 open parentheses negative 2 close parentheses minus 1 end cell row cell limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction end cell equals cell negative 9 end cell end table


Jadi nilai limit as x rightwards arrow negative 2 of fraction numerator 8 x squared plus 14 x minus 4 over denominator 2 x plus 4 end fraction equals negative 9.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Febrianti

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

sama dengan ....

Pembahasan Soal:

Jika limit as x rightwards arrow c of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals 0 over 0, maka 

1. Lakukan pemfaktoran untuk f open parentheses x close parentheses atau g open parentheses x close parentheses , atau

2. Kalikan dengan sekawan pada bentuk akar.

limit as x rightwards arrow 3 of fraction numerator 2 x squared minus 5 x minus 3 over denominator x minus 3 end fraction equals fraction numerator 2.3 squared minus 5.3 minus 3 over denominator 3 minus 3 end fraction equals 0 over 0,

maka dengan pemfaktoran diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 2 x squared minus 5 x minus 3 over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 2 x plus 1 close parentheses open parentheses x minus 3 close parentheses over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses 2 x plus 1 close parentheses end cell row blank equals cell 2.3 plus 1 end cell row blank equals 7 row blank blank blank end table

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 2 x squared minus 5 x minus 3 over denominator x minus 3 end fraction end cell equals 7 row blank blank blank end table.

0

Roboguru

Memfaktorkan

Pembahasan Soal:

Dengan menggunakan metode pemfaktoran diperoleh

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

Dengan demikian nilai table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared minus 1 over denominator x minus 1 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 2 end table.

0

Roboguru

Pembahasan Soal:

Dengan metode pemfaktoran dapat ditentukan hasil dari:

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

Dengan demikian, diperoleh begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 9 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x minus 9 over denominator square root of x minus 3 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 6 end table end style.

0

Roboguru

Tentukan nilai limit dari fungsi berikut : 1) .... (faktorisasi)

Pembahasan Soal:

begin mathsize 14px style lim subscript x rightwards arrow 3 end subscript space x squared plus x minus 5 end style bukanlah merupakan bentuk tak tentu sehingga cara penyelesaiannya bukan dengan pemfaktoran melainkan dapat langsung disubsitusikan dengan begin mathsize 14px style x equals 3 end style, sehingga :

begin mathsize 14px style lim subscript x rightwards arrow 3 end subscript space x squared plus x minus 5 equals 3 squared plus 3 minus 5 equals 7 end style

Dengan demikian, nilai limit tersebut adalah 7

0

Roboguru

Nilai  adalah ....

Pembahasan Soal:

Limit fungsi dengan metode pemfaktoran.

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

begin mathsize 12px style limit as x rightwards arrow negative 2 of space fraction numerator 2 x squared plus 3 x minus 2 over denominator x plus 2 end fraction equals fraction numerator 2 open parentheses negative 2 close parentheses squared plus 3 open parentheses negative 2 close parentheses minus 2 over denominator negative 2 plus 2 end fraction equals fraction numerator 8 minus 6 minus 2 over denominator 4 minus 4 end fraction equals 0 over 0 end style

Karena nilai limit 0 over 0, dan tidak ada bentuk akar maka disederhanakan dengan cara pemfaktoran untuk menentukan nilai limit.

table attributes columnalign right center left columnspacing 2px end attributes row cell limit as x rightwards arrow negative 2 of space fraction numerator 2 x squared plus 3 x minus 2 over denominator x plus 2 end fraction end cell equals cell limit as x rightwards arrow negative 2 of space fraction numerator open parentheses 2 x minus 1 close parentheses up diagonal strike open parentheses x plus 2 close parentheses end strike over denominator up diagonal strike open parentheses x plus 2 close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space open parentheses 2 x minus 1 close parentheses end cell row blank equals cell 2 open parentheses negative 2 close parentheses minus 1 end cell row blank equals cell negative 4 minus 1 end cell row cell limit as x rightwards arrow negative 2 of space fraction numerator 2 x squared plus 3 x minus 2 over denominator x plus 2 end fraction end cell equals cell negative 5 end cell end table

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved