Iklan

Iklan

Pertanyaan

Tentukan nilai limit fungsi berikut. 9. x → 4 lim ​ 3 − 2 x + 1 ​ 8 − 2 x ​

Tentukan nilai limit fungsi berikut.

Iklan

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

nilai dari .

nilai dari begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x over denominator 3 minus square root of 2 x plus 1 end root end fraction equals 6 end style.

Iklan

Pembahasan

Diketahui: Pembagian limit fungsi. Sederhanakan limit fungsi yang diketahui menggunakan metode akar sekawan seperti berikut: Substitusi dengan pada limit fungsi yang telah disederhanakan seperti berikut: Jadi, nilai dari .

Diketahui:

Pembagian limit fungsi.

Sederhanakan limit fungsi yang diketahui menggunakan metode akar sekawan seperti berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x over denominator 3 minus square root of 2 x plus 1 end root end fraction end cell equals cell limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x over denominator 3 minus square root of 2 x plus 1 end root end fraction cross times fraction numerator 3 plus square root of 2 x plus 1 end root over denominator 3 plus square root of 2 x plus 1 end root end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x open parentheses 3 plus square root of 2 x plus 1 end root close parentheses over denominator open parentheses 3 minus square root of 2 x plus 1 end root close parentheses open parentheses 3 plus square root of 2 x plus 1 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x open parentheses 3 plus square root of 2 x plus 1 end root close parentheses over denominator 9 plus 3 square root of 2 x plus 1 end root minus 3 square root of 2 x plus 1 end root minus 2 x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator down diagonal strike 8 minus 2 x end strike open parentheses 3 plus square root of 2 x plus 1 end root close parentheses over denominator down diagonal strike 8 minus 2 x end strike end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space 3 plus square root of 2 x plus 1 end root end cell end table end style

Substitusi begin mathsize 14px style x end style dengan begin mathsize 14px style 4 end style pada limit fungsi yang telah disederhanakan seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space 3 plus square root of 2 x plus 1 end root end cell equals cell 3 plus square root of 2 open parentheses 4 close parentheses plus 1 end root end cell row blank equals cell 3 plus square root of 8 plus 1 end root end cell row blank equals cell 3 plus square root of 9 end cell row blank equals cell 3 plus 3 end cell row blank equals 6 end table end style

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator 8 minus 2 x over denominator 3 minus square root of 2 x plus 1 end root end fraction equals 6 end style.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

3

Iklan

Iklan

Pertanyaan serupa

Tentukan nilai limit berikut. a. ​ ​ lim x → 6 ​ 2 x − 12 3 6 ​ − 3 x ​ ​ ​

1

0.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia