x→0lim​ 1−sec2 2x2x sin 3x​=⋯

Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 0 of space fraction numerator 2 x space sin space 3 x over denominator 1 minus s e c squared space 2 x end fraction equals midline horizontal ellipsis end style 

  1. begin mathsize 14px style 2 over 3 end style 

  2. begin mathsize 14px style 3 over 2 end style 

  3. begin mathsize 14px style negative 2 over 3 end style 

  4. begin mathsize 14px style negative 3 over 2 end style 

  5. begin mathsize 14px style negative 3 over 4 end style 

N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px space fraction numerator size 14px 2 size 14px x size 14px space size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px 1 size 14px minus size 14px s size 14px e size 14px c to the power of size 14px 2 size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px space fraction numerator size 14px 2 size 14px x size 14px space size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px minus size 14px tan to the power of size 14px 2 size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px minus fraction numerator size 14px 2 size 14px x over denominator size 14px tan size 14px space size 14px 2 size 14px x end fraction size 14px times fraction numerator size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px tan size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px minus size 14px 2 over size 14px 2 size 14px times size 14px 3 over size 14px 2 size 14px equals size 14px minus size 14px 3 over size 14px 2 

Maka begin mathsize 14px style limit as x rightwards arrow 0 of space fraction numerator 2 x space sin space 3 x over denominator 1 minus s e c squared space 2 x end fraction equals negative 3 over 2 end style 

40

0.0 (0 rating)

Pertanyaan serupa

Hasil dari  adalah ...

49

5.0

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

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia