Roboguru
SD

x→3lim​ (x−31​−x2−96​)=....

Pertanyaan

limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator x squared minus 9 end fraction close parentheses equals....

  1. 0

  2. 1 over 9

  3. 1 over 6

  4. 1 third

  5. 2 over 3

T. Prita

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jember

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Dalam mengerjakan limit, jika substitusi langsung menghasilkan bentuk tak tentu maka harus menggunakan cara lain dalam mencari nilainya menggunakan operasi aljabar misalnya memfaktorkan, membagi, merasionalkan, dan lainnya.

Beberapa bentuk tak tentu antara lain: 0 over 0 comma space infinity over infinity comma space open parentheses infinity minus infinity close parentheses comma space open parentheses 0 times infinity close parentheses.

Diketahui: 

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator x squared minus 9 end fraction close parentheses end cell end table, dapat dituliskan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator x squared minus 9 end fraction close parentheses end cell equals cell limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator 1 open parentheses x plus 3 close parentheses minus 6 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator x plus 3 minus 6 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator x minus 3 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell end table

Sehingga table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator x squared minus 9 end fraction close parentheses end cell equals cell limit as x rightwards arrow 3 of space fraction numerator x minus 3 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell end table

Substitusi nilai x equals 3 diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of space open parentheses fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 6 over denominator x squared minus 9 end fraction close parentheses end cell equals cell limit as x rightwards arrow 3 of space fraction numerator x minus 3 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 3 minus 3 over denominator open parentheses 3 plus 3 close parentheses open parentheses 3 minus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 3 minus 3 over denominator 6 open parentheses 0 close parentheses end fraction end cell row blank equals cell 0 over 0 space open parentheses text bentuk tak tentu end text close parentheses end cell end table

Selanjutnya dengan memfaktorkan diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of space fraction numerator x minus 3 over denominator x squared minus 9 end fraction end cell equals cell space limit as x rightwards arrow 3 of space fraction numerator x minus 3 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator 1 over denominator x plus 3 end fraction end cell row blank equals cell fraction numerator 1 over denominator 3 plus 3 end fraction end cell row blank equals cell 1 over 6 end cell end table

Oleh karena itu, jawaban yang benar adalah C.

175

4.7 (13 rating)

Antonius J. k

Makasih ❤️

Ilma Nuriyatin

terimakasih

Pertanyaan serupa

y→2lim​y2+y−6y2−4y+4​

86

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