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Nilai  adalah ...

Pertanyaan

Nilai limit as x rightwards arrow 4 of fraction numerator 2 x minus 8 over denominator 16 minus 4 x end fraction adalah ...space 

  1. 1 fourth 

  2. 1 half 

  3. negative 1 half 

  4. negative 1 fourth 

  5. negative 1 over 8 

Pembahasan Soal:

Diketahui limit as x rightwards arrow 4 of fraction numerator 2 x minus 8 over denominator 16 minus 4 x end fraction.

Gunakan metode substitusi langsung.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator 2 x minus 8 over denominator 16 minus 4 x end fraction end cell equals cell fraction numerator 2 times 4 minus 8 over denominator 16 minus 4 times 4 end fraction end cell row blank equals cell 0 over 0 space open parentheses tak space tentu close parentheses end cell end table  

Karena hasilnya tak tentu, maka gunakan metode lain.

Fungsi pada limit merupakan fungsi rasional, maka gunakan metode pemfaktoran untuk menentukan nilai limit.

Nilai dari limit as x rightwards arrow 0 of fraction numerator x squared plus 2 x over denominator x squared minus 3 x end fraction adalah:

 table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator 2 x minus 8 over denominator 16 minus 4 x end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator 2 open parentheses x minus 4 close parentheses over denominator negative 4 open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator 2 over denominator negative 4 end fraction end cell row blank equals cell negative 1 half end cell end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 4 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 x minus 8 over denominator 16 minus 4 x end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table.

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Pembahasan Soal:

Dengan cara pemfaktoran dapat ditentukan nilai limit yang diminta sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 1 of space fraction numerator 2 x squared minus x minus 3 over denominator 3 x squared plus 8 x plus 5 end fraction end cell equals cell limit as x rightwards arrow negative 1 of space fraction numerator open parentheses 2 x minus 3 close parentheses open parentheses x plus 1 close parentheses over denominator open parentheses 3 x plus 5 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of space fraction numerator left parenthesis 2 x minus 3 right parenthesis over denominator left parenthesis 3 x plus 5 right parenthesis end fraction end cell row blank equals cell fraction numerator 2 open parentheses negative 1 close parentheses minus 3 over denominator 3 left parenthesis negative 1 right parenthesis plus 5 end fraction end cell row blank equals cell fraction numerator negative 5 over denominator 2 end fraction end cell row blank equals cell negative 2 1 half end cell end table end style

Dengan demikian, diperoleh begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 x squared minus x minus 3 over denominator 3 x squared plus 8 x plus 5 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table end style.

0

Roboguru

Pembahasan Soal:

Dengan mengalikan akar sekawan, diperoleh perhitungan sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator 1 minus x squared end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator 1 minus x squared end fraction cross times fraction numerator 1 plus square root of x over denominator 1 plus square root of x end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus x over denominator open parentheses 1 minus x squared close parentheses open parentheses 1 plus square root of x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator up diagonal strike 1 minus x end strike over denominator up diagonal strike left parenthesis 1 minus x right parenthesis end strike left parenthesis 1 plus x right parenthesis open parentheses 1 plus square root of x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 over denominator left parenthesis 1 plus x right parenthesis open parentheses 1 plus square root of x close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator left parenthesis 1 plus 1 right parenthesis left parenthesis 1 plus square root of 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 times 2 end fraction end cell row blank equals cell 1 fourth end cell end table end style 

Jadi, diperoleh nilai begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator 1 minus x squared end fraction equals 1 fourth end style.

1

Roboguru

Nilai

Pembahasan Soal:

Dengan menggunakan metode pemfaktoran dan substitusi, nilai limit tersebut yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator 2 x squared plus x minus 3 end fraction end cell equals cell limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator open parentheses 2 x plus 3 close parentheses open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 5 of fraction numerator 1 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 1 over denominator left parenthesis 5 right parenthesis minus 1 end fraction end cell row blank equals cell 1 fourth end cell end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator 2 x squared plus x minus 3 end fraction equals 1 fourth end style

0

Roboguru

Pembahasan Soal:

Diketahui limit as x rightwards arrow negative 1 of fraction numerator x squared minus 4 x minus 5 over denominator x squared plus 7 x plus 6 end fraction, maka:

Gunakan metode subsitusi langsung.

limit as x rightwards arrow negative 1 of fraction numerator x squared minus 4 x minus 5 over denominator x squared plus 7 x plus 6 end fraction equals fraction numerator open parentheses negative 1 close parentheses squared minus 4 times open parentheses negative 1 close parentheses minus 5 over denominator open parentheses negative 1 close parentheses squared plus 7 times open parentheses negative 1 close parentheses plus 6 end fraction equals 0 over 0 space open parentheses tak space tentu close parentheses 

Karena hasilnya tak tentu, dan fungsi limit merupakan fungsi kuadrat, maka gunakan metode pemfaktoran.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 1 of fraction numerator x squared minus 4 x minus 5 over denominator x squared plus 7 x plus 6 end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of fraction numerator open parentheses x minus 5 close parentheses open parentheses x plus 1 close parentheses over denominator open parentheses x plus 6 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of fraction numerator x minus 5 over denominator x plus 6 end fraction end cell row blank equals cell fraction numerator negative 1 minus 5 over denominator negative 1 plus 6 end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 5 end fraction end cell row blank equals cell negative open parentheses 1 1 fifth close parentheses end cell end table 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared minus 4 x minus 5 over denominator x squared plus 7 x plus 6 end fraction end cell end table equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 1 1 fifth close parentheses end cell end table.

0

Roboguru

....

Pembahasan Soal:

Coba substitusikan nilai ke dalam limitnya sehingga diperoleh

x1limx1x31=11(1)31=00

Karena bernilai 00 maka harus dilakukan manipulasi pada limit tersebut dengan metode pemfaktoran. Maka

limx1x1x31====limx1x1(x2+x+1)(x1)limx1x2+x+1(1)2+(1)+13

Dengan demikian, nilai dari limx1x1x31=3.

0

Roboguru

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