Roboguru
SD
SMP
SMA
UTBK/SNBT

Iklan

Iklan

Pertanyaan

Tuliskan dan tunjukkan sifat-sifat limit yang mana saja dapat digunakan untuk menyelesaikan limit fungsi berikut x → 1 lim ​ ( 2 x − 1 ) 4 !

Tuliskan dan tunjukkan sifat-sifat limit yang mana saja dapat digunakan untuk menyelesaikan limit fungsi berikut !

  1.  

  2.  

Ke roboguru plus

Iklan

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Iklan

Pembahasan

Pembahasan
lock

Sifat-sifat limit yang digunakan : 1. 2. 3. 4. 5. Jadi nilai

Sifat-sifat limit yang digunakan :

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

2. begin mathsize 14px style limit as x rightwards arrow a 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 a of f left parenthesis x right parenthesis plus-or-minus limit as x rightwards arrow a of g left parenthesis x right parenthesis end style

3. begin mathsize 14px style limit as x rightwards arrow a of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow a of f left parenthesis x right parenthesis end style

4. begin mathsize 14px style limit as x rightwards arrow a of x equals a end style

5. begin mathsize 14px style limit as x rightwards arrow a of k equals k end style

 

Jadi nilai

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of left parenthesis 2 x minus 1 right parenthesis to the power of 4 end cell equals cell left parenthesis limit as x rightwards arrow 1 of left parenthesis 2 x minus 1 right parenthesis right parenthesis to the power of 4 space end cell row blank equals cell left parenthesis limit as x rightwards arrow 1 of 2 x minus limit as x rightwards arrow 1 of 1 right parenthesis to the power of 4 space end cell row blank equals cell left parenthesis 2 times limit as x rightwards arrow 1 of x minus limit as x rightwards arrow 1 of 1 right parenthesis to the power of 4 space end cell row blank equals cell left parenthesis 2 times open parentheses 1 close parentheses minus 1 right parenthesis to the power of 4 end cell row blank equals cell left parenthesis 2 minus 1 right parenthesis to the power of 4 space end cell row blank equals cell 1 to the power of 4 end cell row blank equals 1 end table end style

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

17

Alfando Eklesio

Ini yang aku cari! Makasih ❤️

Iklan

Iklan

Gold Icon

Klaim Gold gratis sekarang!

Dengan Gold kamu bisa tanya soal ke Forum sepuasnya, lho.

Pertanyaan serupa

x → − 1 lim ​ ( 1 − 2 x ) 3 ....

6

4.0

Jawaban terverifikasi

x → 17 lim ​ x 2 − 64 ​ = . . .

10

4.8

Jawaban terverifikasi

Iklan

Iklan

Jika x → 0 lim ​ f ( x ) = − 5 dan x → 0 lim ​ g ( x ) = 1 , maka x → 0 lim ​ ( g 2 ( x ) f 2 ( x ) ​ + f ( x ) g ( x ) ​ + 9 ​ ) = ...

1

3.0

Jawaban terverifikasi

x → 2 lim ​ x 2 + 3 x = …

3

5.0

Jawaban terverifikasi

Hitunglah nilai x → − 1 lim ​ ( x 2 − 3 x + 2 )

9

0.0

Jawaban terverifikasi
Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Download di Google Play

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Bantuan & Panduan

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