Roboguru

limit as x rightwards arrow 3 of space fraction numerator sin space left parenthesis 2 x minus 6 right parenthesis over denominator square root of 4 minus x end root minus 1 end fraction equals space..

  1. 4

  2. 2

  3. 0

  4. -2

  5. -4

Jawaban:

Gunakan space dalil space straight L space hospital  limit as straight x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 straight x minus 6 close parentheses over denominator square root of 4 minus straight x end root minus 1 end fraction  equals limit as straight x rightwards arrow 3 of invisible function application fraction numerator 2 cos invisible function application open parentheses 2 straight x minus 6 close parentheses over denominator negative fraction numerator 1 over denominator 2 square root of 4 minus straight x end root end fraction end fraction  equals fraction numerator 2 cos invisible function application open parentheses 2 open parentheses 3 close parentheses minus 6 close parentheses over denominator negative fraction numerator 1 over denominator 2 square root of 4 minus 3 end root end fraction end fraction  equals 2 cos invisible function application 0. open parentheses negative 2 close parentheses  equals box enclose negative 4 end enclose

0

Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved