Iklan

Pertanyaan

x → 3 lim ​ x − 4 x − 3 ​ 5 x − 15 ​ = …

  1. begin mathsize 14px style horizontal ellipsis end style

  2. begin mathsize 14px style horizontal ellipsis end style

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

02

:

12

:

02

:

38

Klaim

Iklan

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

.

 begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction equals 15 end style.

Pembahasan

Jika kita melakukan substitusi langsung pada permasalahan limit di atas, kita akan memperoleh bentuk tak tentu . Oleh karena itu, kita akan menyelesaikan masalah limit tersebut dengan cara mengalikan akar sekawan penyebutnya. Perhatikan bahwa Dengan demikian, .

Jika kita melakukan substitusi langsung pada permasalahan limit di atas, kita akan memperoleh bentuk tak tentu begin mathsize 14px style 0 over 0 end style. Oleh karena itu, kita akan menyelesaikan masalah limit tersebut dengan cara mengalikan akar sekawan penyebutnya.

Perhatikan bahwa

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 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction times fraction numerator x plus square root of 4 x minus 3 end root over denominator x plus square root of 4 x minus 3 end root end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus open parentheses square root of 4 x minus 3 end root close parentheses squared end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus open parentheses 4 x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus 4 x plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 5 up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 5 open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus square root of 4 times 3 minus 3 end root close parentheses over denominator open parentheses 3 minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus square root of 9 close parentheses over denominator open parentheses 2 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus 3 close parentheses over denominator 2 end fraction end cell row blank equals 15 end table end style

Dengan demikian, begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction equals 15 end style.

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

21

Iklan

Pertanyaan serupa

lim x → 2 ​ 3 − x 2 + 5 ​ ( 4 − x 2 ) ​ ​ = ​ .... ​

3

4.8

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

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

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