Roboguru

Nilai dari  adalah ...

Pertanyaan

Nilai dari limit as x rightwards arrow infinity of space open parentheses square root of x squared plus 4 x end root minus x close parentheses adalah ...

Pembahasan Soal:

Diketahui limit as x rightwards arrow infinity of space open parentheses square root of x squared plus 4 x end root minus x close parentheses.

Nilai dari limit as x rightwards arrow infinity of space open parentheses square root of x squared plus 4 x end root minus x close parentheses:

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

Jadi, nilai dari limit as x rightwards arrow infinity of space open parentheses square root of x squared plus 4 x end root minus x close parentheses adalah 2

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Dwi

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 02 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

x→∞lim​3x2−x−2​−5x2+2x−3​=...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow infinity of square root of ax squared plus bx plus straight c end root minus square root of px squared plus qx plus straight r end root end cell row blank blank cell open curly brackets table row cell straight a greater than straight p rightwards double arrow plus infinity end cell row cell straight a equals straight p rightwards double arrow fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row cell straight a less than straight p rightwards double arrow negative infinity end cell end table close end cell row blank blank blank row blank blank cell limit as straight x rightwards arrow infinity of square root of 3 straight x squared minus straight x minus 2 end root minus square root of 5 straight x squared plus 2 straight x minus 3 end root end cell row 3 less than cell 5 comma space straight a less than straight p rightwards double arrow negative infinity end cell end table end style 

Jadi, xlim3x2x25x2+2x3=

0

Roboguru

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow infinity of square root of ax squared plus bx plus straight c end root minus square root of px squared plus qx plus straight r end root end cell row straight a equals cell straight p rightwards double arrow fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row blank blank blank row blank blank cell limit as straight x rightwards arrow infinity of square root of 4 straight x squared minus straight x plus 7 end root minus left parenthesis 2 straight x plus 1 right parenthesis end cell row blank equals cell limit as straight x rightwards arrow infinity of square root of 4 straight x squared minus straight x plus 7 end root minus square root of left parenthesis 2 straight x plus 1 right parenthesis squared end root end cell row blank equals cell limit as straight x rightwards arrow infinity of square root of 4 straight x squared minus straight x plus 7 end root minus square root of 4 straight x squared plus 4 straight x plus 1 end root end cell row blank equals cell fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row blank equals cell fraction numerator negative 1 minus 4 over denominator 2 square root of 4 end fraction end cell row blank equals cell fraction numerator negative 5 over denominator 2 times 2 end fraction end cell row blank equals cell negative 5 over 4 end cell end table end style 

0

Roboguru

Nilai ....

Pembahasan Soal:

Pada limit fungsi aljabar bentuk akar, pabila menggunkan metode substitusi menghasilkan nilai limit begin mathsize 14px style 0 over 0 end style, maka dapat diterapkan metode kali akar sekawan sebagai berikut

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack l i m with x rightwards arrow negative 1 below fraction numerator x squared minus 1 over denominator square root of 8 minus x end root minus 3 end fraction end cell equals cell stack l i m with x rightwards arrow negative 1 below fraction numerator x squared minus 1 over denominator square root of 8 minus x end root minus 3 end fraction fraction numerator square root of 8 minus x end root plus 3 over denominator square root of 8 minus x end root plus 3 end fraction end cell row blank equals cell stack l i m with x rightwards arrow negative 1 below fraction numerator open parentheses x squared minus 1 close parentheses open parentheses square root of 8 minus x end root plus 3 close parentheses over denominator 8 minus x minus 9 end fraction end cell row blank equals cell stack l i m with x rightwards arrow negative 1 below fraction numerator open parentheses x minus 1 close parentheses left parenthesis x plus 1 right parenthesis open parentheses square root of 8 minus x end root plus 3 close parentheses over denominator negative left parenthesis x plus 1 right parenthesis end fraction end cell row blank equals cell stack l i m with x rightwards arrow negative 1 below minus open parentheses x minus 1 close parentheses open parentheses square root of 8 minus x end root plus 3 close parentheses end cell row blank equals cell negative left parenthesis negative 1 minus 1 right parenthesis open parentheses square root of 8 minus left parenthesis negative 1 right parenthesis end root plus 3 close parentheses end cell row blank equals cell 2 open parentheses square root of 9 plus 3 close parentheses end cell row blank equals cell 2 cross times 6 end cell row blank equals 12 end table end style

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow infinity of square root of ax squared plus bx plus straight c end root minus square root of px squared plus qx plus straight r end root end cell row blank blank cell open curly brackets table row cell straight a greater than straight p rightwards double arrow plus infinity end cell row cell straight a equals straight p rightwards double arrow fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row cell straight a less than straight p rightwards double arrow negative infinity end cell end table close end cell row blank blank blank row blank blank cell limit as straight x rightwards arrow infinity of square root of 4 straight x squared plus 8 straight x plus 1 end root minus square root of straight x squared plus 3 straight x minus 4 end root end cell row 4 greater than cell 1 comma straight a greater than straight p rightwards double arrow plus infinity end cell end table end style 

0

Roboguru

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow infinity of square root of ax squared plus bx plus straight c end root minus square root of px squared plus qx plus straight r end root end cell row straight a equals cell straight p rightwards double arrow fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row blank blank blank row blank blank cell limit as straight x rightwards arrow infinity of left parenthesis 5 straight x plus 1 right parenthesis minus square root of 25 straight x squared minus 10 straight x plus 4 end root end cell row blank equals cell limit as straight x rightwards arrow infinity of square root of left parenthesis 5 straight x plus 1 right parenthesis squared end root minus square root of 25 straight x squared minus 10 straight x plus 4 end root end cell row blank equals cell limit as straight x rightwards arrow infinity of square root of 25 straight x squared plus 10 straight x plus 1 end root minus square root of 25 straight x squared minus 10 straight x plus 4 end root end cell row blank equals cell fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row blank equals cell fraction numerator 10 minus left parenthesis negative 10 right parenthesis over denominator 2 square root of 25 end fraction end cell row blank equals cell fraction numerator 10 plus 10 over denominator 2 times 5 end fraction end cell row blank equals cell 20 over 10 end cell row blank equals 2 end table end style 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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