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Nilai dari

Pertanyaan

Nilai dari limit as x rightwards arrow infinity of square root of 4 x squared minus 4 x plus 3 end root minus square root of 4 x squared plus 4 x plus 3 end root equals...

  1. negative 2

  2. negative 1

  3. 0

  4. 1

  5. negative 2

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of square root of a x squared plus b x plus 3 end root minus square root of p x squared plus q x plus r end root end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction comma space untuk space a end cell equals p row blank blank cell limit as x rightwards arrow infinity of square root of 4 x squared minus 4 x plus 3 end root minus square root of 4 x squared plus 4 x plus 3 end root end cell row a equals cell p equals 4 end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell equals cell fraction numerator negative 4 minus 4 over denominator 2 square root of 4 end fraction end cell row blank equals cell fraction numerator negative 8 over denominator 4 end fraction end cell row blank equals cell negative 2 end cell row blank blank blank end table

Jadi, nilai dari limit as x rightwards arrow infinity of square root of 4 x squared minus 4 x plus 3 end root minus square root of 4 x squared plus 4 x plus 3 end root equals negative 2.

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Herlanda

Mahasiswa/Alumni STKIP PGRI Jombang

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Nilai dari adalah...

Pembahasan Soal:

Ingat kembali cara cepat untuk menentukan nilai imit tak hingga bentuk akar berikut.

xlimax2+bx+cax2+px+q=2abp

Sehingga semua bentuk pada soal harus diubah menjadi bentuk akar.

Perhatikan perhitungan berikut.

========limx4x23x+7(2x3)limx4x23x+7(2x3)2limx4x23x+7(2x3)(2x3)limx4x23x+74x26x6x+9limx4x23x+74x212x+9243(12)223+1249241 

Sehingga diperoleh nilai xlim4x23x+7(2x3)=241.

Oleh karena itu, jawaban yang benar adalah D.

Roboguru

Pembahasan Soal:

begin mathsize 14px style 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 open curly brackets table row cell straight a greater than straight p rightwards arrow plus infinity end cell row cell straight a equals straight p rightwards 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 arrow negative infinity end cell end table close  limit as straight x rightwards arrow infinity of open parentheses 4 straight x minus 2 close parentheses minus square root of 4 straight x squared minus 8 straight x plus 1 end root equals limit as straight x rightwards arrow infinity of square root of open parentheses 4 straight x minus 2 close parentheses squared end root minus square root of 4 straight x squared minus 8 straight x plus 1 end root equals limit as straight x rightwards arrow infinity of square root of 16 straight x squared minus 16 straight x plus 4 end root minus square root of 4 straight x squared minus 8 straight x plus 1 end root 16 greater than 4 rightwards double arrow straight a greater than straight p rightwards double arrow plus infinity end style  

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.

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 

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.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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