Iklan

Iklan

Pertanyaan

x → 1 lim ​ sin 2 ( 2 x − 2 ) ( x 2 − 1 ) tan 2 ( x − 1 ) ​

  

Iklan

H. Endah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Jawaban terverifikasi

Iklan

Pembahasan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator left parenthesis x squared minus 1 right parenthesis space tan space 2 left parenthesis x minus 1 right parenthesis over denominator sin squared space left parenthesis 2 x minus 2 right parenthesis end fraction end cell equals cell limit as x rightwards arrow 1 of space fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses space tan space 2 left parenthesis x minus 1 right parenthesis over denominator sin space 2 open parentheses x minus 1 close parentheses space times space sin space 2 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator left parenthesis x plus 1 right parenthesis left parenthesis x minus 1 right parenthesis space times space 2 left parenthesis x minus 1 right parenthesis over denominator 2 left parenthesis x minus 1 right parenthesis space times space 2 left parenthesis x minus 1 right parenthesis end fraction space times space fraction numerator space tan space 2 left parenthesis x minus 1 right parenthesis over denominator 2 open parentheses x minus 1 close parentheses end fraction space times space fraction numerator 2 left parenthesis x minus 1 right parenthesis over denominator sin space 2 left parenthesis x minus 1 right parenthesis end fraction space times space fraction numerator 2 left parenthesis x minus 1 right parenthesis over denominator sin space 2 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator left parenthesis x plus 1 right parenthesis over denominator 2 end fraction space times space fraction numerator space tan space 2 left parenthesis x minus 1 right parenthesis over denominator 2 left parenthesis x minus 1 right parenthesis end fraction space times space fraction numerator 2 left parenthesis x minus 1 right parenthesis over denominator sin space 2 left parenthesis x minus 1 right parenthesis end fraction space times space fraction numerator 2 left parenthesis x minus 1 right parenthesis over denominator sin space 2 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell fraction numerator left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction space times space 1 space times space 1 space times space 1 end cell row blank equals cell 2 over 2 end cell row blank equals 1 end table end style 

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

5

Alysha

Pembahasan terpotong

Iklan

Iklan

Pertanyaan serupa

Nilai x → 3 π ​ lim ​ sin ( 3 x − π ) tan ( 3 x − π ) cos ( 2 x ) ​ adalah ...

5

4.1

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