Iklan

Iklan

Pertanyaan

x → 0 lim ​ sin 2 x − sin x sin 3 x + sin x ​ = ...

                    

  1.  4    

  2. 2        

  3. 4 over 3  

  4. 1             

  5. 0  

Iklan

P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

jawaban yang benar adalah A.

Iklan

Pembahasan

Ingatlah sifat-sifat limit fungsi: Dengan menggunakan sifat-sifat limit fungsi tersebut, didapatkan : Maka, . Oleh karena itu, jawaban yang benar adalah A.

Ingatlah sifat-sifat limit fungsi:

  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis plus g left parenthesis x right parenthesis close square brackets equals 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
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis close square brackets equals 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
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis divided by g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis divided by limit as x rightwards arrow c of g left parenthesis x right parenthesis
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals a over b

Dengan menggunakan sifat-sifat limit fungsi tersebut, didapatkan :

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator sin space 3 x plus sin space x over denominator sin space 2 x minus sin space x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses sin space 3 x plus sin space x close parentheses over denominator open parentheses sin space 2 x minus sin space x close parentheses end fraction cross times fraction numerator 1 over x over denominator 1 over x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space open parentheses fraction numerator begin display style fraction numerator sin space 3 x over denominator x end fraction plus fraction numerator sin space x over denominator x end fraction end style over denominator begin display style fraction numerator sin space 2 x over denominator x end fraction minus fraction numerator sin space x over denominator x end fraction end style end fraction close parentheses end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space 3 x over denominator x end fraction plus fraction numerator sin space x over denominator x end fraction close parentheses end style over denominator begin display style limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space 2 x over denominator x end fraction minus fraction numerator sin space x over denominator x end fraction close parentheses end style end fraction end cell row blank equals cell fraction numerator limit as x rightwards arrow 0 of space begin display style fraction numerator sin space 3 x over denominator x end fraction end style plus limit as x rightwards arrow 0 of space begin display style fraction numerator sin space x over denominator x end fraction end style over denominator limit as x rightwards arrow 0 of space begin display style fraction numerator sin space 2 x over denominator x end fraction end style minus limit as x rightwards arrow 0 of begin display style fraction numerator sin space x over denominator x end fraction end style space end fraction end cell row blank equals cell fraction numerator 3 plus 1 over denominator 2 minus 1 end fraction end cell row blank equals cell 4 over 1 end cell row blank equals 4 end table     

Maka, limit as x rightwards arrow 0 of space fraction numerator sin space 3 x plus sin space x over denominator sin space 2 x minus sin space x end fraction equals 1

Oleh karena itu, jawaban yang benar adalah A.

Latihan Bab

Konsep Kilat

Limit Fungsi Trigonometri

Limit Tak Hingga

Kekontinuan dan Asimtot

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

85

Iklan

Iklan

Pertanyaan serupa

x → 0 lim ​ x + sin 5 x tan 2 x + sin x ​ = ...

58

5.0

Jawaban terverifikasi

Iklan

Iklan

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

Fitur Roboguru

Topik Roboguru

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2023 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia