Iklan

Iklan

Pertanyaan

Jika L , K adalah bilangan real dan x → c lim ​ f ( x ) = L , x → c lim ​ g ( x ) = K maka tentukan x → c lim ​ ( f ( x ) + g ( x ) f ( x ) − g ( x ) ​ ) 2 !

Jika ,  adalah bilangan real dan ,  maka tentukan !

  1. undefined 

  2. undefined 

Iklan

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Jawaban

.

 size 14px lim with size 14px x size 14px rightwards arrow size 14px c below begin mathsize 14px style left parenthesis fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction right parenthesis end style to the power of size 14px 2 size 14px equals fraction numerator begin mathsize 14px style left parenthesis L minus K right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis L plus K right parenthesis squared end style end fraction.

Iklan

Pembahasan

Pembahasan
lock

Sifat yang digunakan: 1. ; 2. ; dan 3. . Dapat diperoleh, Jadi, .

Sifat yang digunakan:

1. begin mathsize 14px style limit as x rightwards arrow c of open parentheses f left parenthesis x right parenthesis close parentheses to the power of n equals open parentheses limit as x rightwards arrow c of f left parenthesis x right parenthesis close parentheses to the power of n end style;

2. undefined; dan

3. begin mathsize 14px style limit as x rightwards arrow c of open parentheses f left parenthesis x right parenthesis plus-or-minus g left parenthesis x right parenthesis close parentheses equals limit as x rightwards arrow c of f left parenthesis x right parenthesis end style.

 

Dapat diperoleh,

begin mathsize 14px style limit as x rightwards arrow c of open parentheses fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses limit as x rightwards arrow c of fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses fraction numerator limit as x rightwards arrow c of open parentheses f open parentheses x close parentheses minus g left parenthesis x right parenthesis close parentheses over denominator limit as x rightwards arrow c of open parentheses f open parentheses x close parentheses plus g left parenthesis x right parenthesis close parentheses end fraction close parentheses squared equals open parentheses fraction numerator limit as x rightwards arrow c of f left parenthesis x right parenthesis minus limit as x rightwards arrow c of g left parenthesis x right parenthesis over denominator limit as x rightwards arrow c of f left parenthesis x right parenthesis plus limit as x rightwards arrow c of g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses fraction numerator L minus K over denominator L plus K end fraction close parentheses squared equals open parentheses L minus K close parentheses squared over open parentheses L plus K close parentheses squared end style  

Jadi, size 14px lim with size 14px x size 14px rightwards arrow size 14px c below begin mathsize 14px style left parenthesis fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction right parenthesis end style to the power of size 14px 2 size 14px equals fraction numerator begin mathsize 14px style left parenthesis L minus K right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis L plus K right parenthesis squared end style end fraction.

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

38

Iklan

Iklan

Pertanyaan serupa

Tentukan limit fungsi dari: x → 0 lim ​ x − 2 4 x 2 + 1 ​ = ....

43

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