Roboguru

Nilai

Pertanyaan

Nilai limit as x rightwards arrow 4 of space fraction numerator 20 x squared minus 320 over denominator x plus square root of x minus 6 end fraction equals horizontal ellipsis 

Pembahasan Soal:

Nilainya didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator 20 x squared minus 320 over denominator x plus square root of x minus 6 end fraction end cell equals cell limit as x rightwards arrow 4 of space fraction numerator 20 x squared minus 320 over denominator open parentheses x minus 6 close parentheses plus square root of x end fraction cross times fraction numerator open parentheses x minus 6 close parentheses minus square root of x over denominator open parentheses x minus 6 close parentheses minus square root of x end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator open parentheses 20 x squared minus 320 close parentheses open parentheses open parentheses x minus 6 close parentheses minus square root of x close parentheses over denominator open parentheses x minus 6 close parentheses squared plus open parentheses square root of x close parentheses squared end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator open parentheses 20 x squared minus 320 close parentheses open parentheses open parentheses x minus 6 close parentheses minus square root of x close parentheses over denominator x squared minus 12 x plus 36 minus x end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 20 open parentheses x squared minus 16 close parentheses open parentheses open parentheses x minus 6 close parentheses minus square root of x close parentheses over denominator x squared minus 13 x plus 36 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 20 open parentheses x minus 4 close parentheses open parentheses x plus 4 close parentheses open parentheses open parentheses x minus 6 close parentheses minus square root of x close parentheses over denominator open parentheses x minus 4 close parentheses open parentheses x minus 9 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 20 open parentheses x plus 4 close parentheses open parentheses open parentheses x minus 6 close parentheses minus square root of x close parentheses over denominator open parentheses x minus 9 close parentheses end fraction end cell row blank equals cell fraction numerator 20 open parentheses 4 plus 4 close parentheses open parentheses open parentheses 4 minus 6 close parentheses minus square root of 4 close parentheses over denominator open parentheses 4 minus 9 close parentheses end fraction end cell row blank equals cell fraction numerator negative 640 over denominator negative 5 end fraction end cell row blank equals 128 end table

Jadi, nilai limit as x rightwards arrow 4 of space fraction numerator 20 x squared minus 320 over denominator x plus square root of x minus 6 end fraction equals 128

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Endah

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai  adalah ....

Pembahasan Soal:

Dengan mengalikan akar sekawan.

begin mathsize 14px style limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction times fraction numerator square root of 12 plus square root of x plus 8 end root over denominator square root of 12 plus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator 12 minus open parentheses x plus 8 close parentheses end fraction equals limit as x rightwards arrow 4 of fraction numerator up diagonal strike open parentheses 4 minus x close parentheses end strike open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator up diagonal strike open parentheses 4 minus x close parentheses end strike end fraction equals limit as x rightwards arrow 4 of square root of 12 plus square root of x plus 8 end root equals square root of 12 plus square root of 4 plus 8 end root equals square root of 12 plus square root of 12 equals 2 square root of 12 equals 2 square root of 4 times 3 end root equals 2 times 2 square root of 3 equals 4 square root of 3 end style 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Hitunglah setiap limit fungsi berikut. b.

Pembahasan Soal:

Untuk menemukan nilai limitnya maka dikalikan dengan akar sekawan dari penyebutnyaspace 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as a rightwards arrow b of fraction numerator a square root of a minus b square root of b over denominator square root of a minus square root of b end fraction end cell equals cell limit as a rightwards arrow b of fraction numerator a square root of a minus b square root of b over denominator square root of a minus square root of b end fraction times fraction numerator square root of a plus square root of b over denominator square root of a plus square root of b end fraction end cell row blank equals cell limit as a rightwards arrow b of fraction numerator a squared plus a square root of a b end root minus b square root of a b end root minus b squared over denominator a minus b end fraction end cell row blank equals cell limit as a rightwards arrow b of fraction numerator a squared minus b squared plus square root of a b end root open parentheses a minus b close parentheses over denominator a minus b end fraction end cell row blank equals cell limit as a rightwards arrow b of fraction numerator a squared minus b squared over denominator a minus b end fraction plus fraction numerator square root of a b end root left parenthesis a minus b right parenthesis over denominator a minus b end fraction space space space space space comma bagi space menjadi space dua space bentuk space pecahan end cell row blank equals cell limit as a rightwards arrow b of fraction numerator up diagonal strike left parenthesis a minus b right parenthesis end strike left parenthesis a plus b right parenthesis over denominator up diagonal strike left parenthesis a minus b right parenthesis end strike end fraction plus fraction numerator square root of a b end root up diagonal strike left parenthesis a minus b right parenthesis end strike over denominator up diagonal strike left parenthesis a minus b right parenthesis end strike end fraction end cell row blank equals cell limit as a rightwards arrow b of left parenthesis a plus b right parenthesis plus square root of a b end root end cell row blank equals cell open parentheses b plus b close parentheses plus square root of b squared end root space space space comma space substitusi space nilai space a equals b end cell row blank equals cell 2 b plus b end cell row blank equals cell 3 b end cell end table end style

0

Roboguru

.

Pembahasan Soal:

Limit fungsi dengan metode mengalikan akar sekawan.

Cek limit terlebih dahulu jika nilai limit 0 over 0 maka harus disederhanakan dengan pemfaktoran atau mengalikan dengan akar sekawan.

limit as x rightwards arrow 3 of space fraction numerator open parentheses square root of x minus square root of 3 close parentheses over denominator x minus 3 end fraction equals fraction numerator open parentheses square root of 3 minus square root of 3 close parentheses over denominator 3 minus 3 end fraction equals 0 over 0

Karena nilai limit 0 over 0, dan terdapat bentuk akar maka disederhanakan dengan cara mengalikan dengan akar sekawan.

table attributes columnalign right center left columnspacing 2px end attributes row cell limit as x rightwards arrow 3 of space fraction numerator open parentheses square root of x minus square root of 3 close parentheses over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses square root of x minus square root of 3 close parentheses over denominator x minus 3 end fraction cross times fraction numerator square root of x plus square root of 3 over denominator square root of x plus square root of 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator up diagonal strike open parentheses x minus 3 close parentheses end strike over denominator up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses square root of x plus square root of 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator 1 over denominator open parentheses square root of x plus square root of 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 3 plus square root of 3 end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell fraction numerator square root of 3 over denominator 2 cross times 3 end fraction end cell row cell limit as x rightwards arrow 3 of space fraction numerator open parentheses square root of x minus square root of 3 close parentheses over denominator x minus 3 end fraction end cell equals cell 1 over 6 square root of 3 end cell end table

Jadi, diperoleh limit as x rightwards arrow 3 of space fraction numerator open parentheses square root of x minus square root of 3 close parentheses over denominator x minus 3 end fraction equals 1 over 6 square root of 3.

0

Roboguru

Nilai dari

Pembahasan Soal:

Nilainya didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction end cell equals cell limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction times fraction numerator square root of x plus 12 end root plus 3 over denominator square root of x plus 12 end root plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator x plus 12 minus 9 over denominator open parentheses x squared plus 7 x plus 12 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator up diagonal strike x plus 3 end strike over denominator up diagonal strike open parentheses x plus 3 close parentheses end strike open parentheses x plus 4 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator 1 over denominator open parentheses x plus 4 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses open parentheses negative 3 close parentheses plus 4 close parentheses open parentheses square root of open parentheses negative 3 close parentheses plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses 3 plus 3 close parentheses end fraction end cell row blank equals cell 1 over 6 end cell end table  

Dengan demikian, nilai dari limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction equals 1 over 6.

Jadi, jawaban yang benar adalah D.

0

Roboguru

Nilai

Pembahasan Soal:

Untuk menyelesaikan fungsi limit di atas dapat menggunakan konsep mengalikan dengan akar sekawan, sehinga:

limit as x rightwards arrow 6 of fraction numerator square root of 3 x minus 2 end root minus square root of 2 x plus 4 end root over denominator x minus 6 end fraction equals

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 6 of fraction numerator square root of 3 x minus 2 end root minus square root of 2 x plus 4 end root over denominator x minus 6 end fraction cross times fraction numerator square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root over denominator square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root end fraction end cell row blank equals cell limit as x rightwards arrow 6 of fraction numerator open parentheses square root of 3 x minus 2 end root minus square root of 2 x plus 4 end root close parentheses open parentheses square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root close parentheses over denominator left parenthesis x minus 6 right parenthesis open parentheses square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 6 of fraction numerator 3 x minus 2 minus open parentheses 2 x plus 4 close parentheses over denominator left parenthesis x minus 6 right parenthesis open parentheses square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 6 of fraction numerator x minus 6 over denominator left parenthesis x minus 6 right parenthesis open parentheses square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 6 of fraction numerator 1 over denominator open parentheses square root of 3 x minus 2 end root plus square root of 2 x plus 4 end root close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses square root of 3 open parentheses 6 close parentheses minus 2 end root plus square root of 2 open parentheses 6 close parentheses plus 4 end root close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 16 plus square root of 16 end fraction end cell row blank equals cell 1 over 8 end cell end table

Jadi, Nilai limit as x rightwards arrow 6 of fraction numerator square root of 3 x minus 2 end root minus square root of 2 x plus 4 end root over denominator x minus 6 end fraction adalah 1 over 8 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

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