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Tentukan nilai dari setiap limit fungsi berikut dengan metode pemfaktoran!

Pertanyaan

Tentukan nilai dari setiap limit fungsi berikut dengan metode pemfaktoran!

 limit as x rightwards arrow 2 of fraction numerator x squared minus 5 x plus 6 over denominator x squared minus x minus 2 end fraction        

Pembahasan Soal:

limit as x rightwards arrow 2 of fraction numerator x squared minus 5 x plus 6 over denominator x squared minus x minus 2 end fraction 

Subtitusikan x equals 2 ke persamaan limit tersebut, sehingga diperoleh  sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x squared minus 5 x plus 6 over denominator x squared minus x minus 2 end fraction end cell equals cell fraction numerator 2 squared minus 5 open parentheses 2 close parentheses plus 6 over denominator 2 squared minus 2 minus 2 end fraction end cell row blank equals cell fraction numerator 4 minus 10 plus 6 over denominator 4 minus 2 minus 2 end fraction end cell row blank equals cell 0 over 0 end cell end table  

Karena menghasilkan limit bentuk tak tentu, maka nilai limit tersebut dapat ditentukan dengan metode pemfaktoran berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x squared minus 5 x plus 6 over denominator x squared minus x minus 2 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator open parentheses x minus 3 close parentheses up diagonal strike open parentheses x minus 2 close parentheses end strike over denominator open parentheses x plus 1 close parentheses up diagonal strike open parentheses x minus 2 close parentheses end strike end fraction end cell row blank equals cell fraction numerator 2 minus 3 over denominator 2 plus 1 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 3 end fraction end cell end table      

   Jadi,  Nilai dari limit as x rightwards arrow 2 of fraction numerator x squared minus 5 x plus 6 over denominator x squared minus x minus 2 end fraction adalah negative 1 third

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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Pembahasan Soal:

Gunakan pemfaktoran untuk menyederhanakan fungsi limit.

begin mathsize 14px style table attributes columnalign center center left end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x squared plus 5 x minus 14 over denominator x squared plus 11 x plus 28 end fraction end cell equals cell limit as x rightwards arrow negative 7 of space fraction numerator x squared plus 7 x minus 2 x minus 14 over denominator x squared plus 7 x plus 4 x plus 28 end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator x left parenthesis x plus 7 right parenthesis minus 2 left parenthesis x plus 7 right parenthesis over denominator x left parenthesis x plus 7 right parenthesis plus 4 left parenthesis x plus 7 right parenthesis end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator down diagonal strike left parenthesis x plus 7 right parenthesis end strike left parenthesis x minus 2 right parenthesis over denominator down diagonal strike left parenthesis x plus 7 right parenthesis end strike left parenthesis x plus 4 right parenthesis end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator x minus 2 over denominator x plus 4 end fraction end cell row space space space row space space space row space space cell space space end cell end table end style 

Kemudian substitusi begin mathsize 14px style x equals negative 7 end style ke seperti berikut:

begin mathsize 14px style table attributes columnalign center center left end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x minus 2 over denominator x plus 4 end fraction end cell equals cell fraction numerator open parentheses negative 7 close parentheses minus 2 over denominator open parentheses negative 7 close parentheses plus 4 end fraction end cell row space equals cell fraction numerator negative 9 over denominator negative 3 end fraction end cell row space equals 3 end table end style 

 

Roboguru

Hitunglah limit fungsi berikut!

Pembahasan Soal:

begin mathsize 14px style x equals 5 end style disubtitusikan pada limit tersebut

   begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator x squared minus 25 over denominator x minus 5 end fraction equals fraction numerator 5 squared minus 25 over denominator 5 minus 5 end fraction equals 0 over 0 end style 

Karena menghasilkan limit bentuk tak tentu, maka akan ditentukan dengan metode pemfaktoran

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 x squared minus 25 over denominator x minus 5 end fraction end cell equals cell fraction numerator up diagonal strike open parentheses x minus 5 close parentheses end strike open parentheses x plus 5 close parentheses over denominator up diagonal strike open parentheses x minus 5 close parentheses end strike end fraction end cell row cell space space space space space space space space space space space space space space space space space space end cell equals cell 5 plus 5 end cell row cell space space space space space space space space space space space space space space space space space space end cell equals 10 end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator x squared minus 25 over denominator x minus 5 end fraction end style adalah 10  

Roboguru

Tentukan nilai limit fungsi berikut.

Pembahasan Soal:

Diketahui:

Pembagian limit fungsi.

Sederhanakan limit fungsi yang diketahui menggunakan metode pemfaktoran seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator x squared plus 5 x minus 36 over denominator x squared minus 3 x minus 4 end fraction end cell equals cell limit as x rightwards arrow 4 of space fraction numerator x squared plus 9 x minus 4 x minus 36 over denominator x squared plus x minus 4 x minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator x open parentheses x plus 9 close parentheses minus 4 open parentheses x plus 9 close parentheses over denominator x open parentheses x plus 1 close parentheses minus 4 open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator open parentheses x plus 9 close parentheses down diagonal strike open parentheses x minus 4 close parentheses end strike over denominator open parentheses x plus 1 close parentheses down diagonal strike open parentheses x minus 4 close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator x plus 9 over denominator x plus 1 end fraction end cell end table end style

Substitusi begin mathsize 14px style x end style dengan begin mathsize 14px style 4 end style pada limit fungsi yang telah disederhanakan seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator x plus 9 over denominator x plus 1 end fraction end cell equals cell fraction numerator 4 plus 9 over denominator 4 plus 1 end fraction end cell row blank equals cell 13 over 5 end cell end table end style

Maka, nilai dari begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator x squared plus 5 x minus 36 over denominator x squared minus 3 x minus 4 end fraction equals 13 over 5 end style.

Roboguru

Selesaikan setiap masalah limit di bawah ini secara intuitif.

Pembahasan Soal:

Jika limit pada soal di atas langsung disubstitusikan dengan nilai x equals negative 2, maka akan memberikan hasil 0 over 0. Sehingga, cara ini tidak dapat dilakukan pada soal tersebut.

Dengan menggunakan konsep limit dan metode pemfaktoran, maka nilai dari limit as x rightwards arrow negative 2 of open parentheses fraction numerator 3 x squared plus 5 x minus 2 over denominator x plus 2 end fraction close parentheses adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of open parentheses fraction numerator 3 x squared plus 5 x minus 2 over denominator x plus 2 end fraction close parentheses end cell equals cell limit as x rightwards arrow negative 2 of space fraction numerator open parentheses 3 x squared plus 5 x minus 2 close parentheses over denominator open parentheses x plus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space fraction numerator up diagonal strike open parentheses x plus 2 close parentheses end strike open parentheses 3 x minus 1 close parentheses over denominator up diagonal strike open parentheses x plus 2 close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of space open parentheses 3 x minus 1 close parentheses end cell row blank equals cell 3 times open parentheses negative 2 close parentheses minus 1 end cell row blank equals cell negative 6 minus 1 end cell row cell limit as x rightwards arrow negative 2 of open parentheses fraction numerator 3 x squared plus 5 x minus 2 over denominator x plus 2 end fraction close parentheses end cell equals cell negative 7 end cell end table

Dengan demikian, nilai dari limit as x rightwards arrow negative 2 of open parentheses fraction numerator 3 x squared plus 5 x minus 2 over denominator x plus 2 end fraction close parentheses adalah negative 7. 

Roboguru

....

Pembahasan Soal:

Substitusi limit dengan x=3, diperoleh

x3x2+4x21===3332+4(3)2109+122100

Ingat bahwa, suatu limit dengan metode substitusi yang menghasilkan nilai limit begin mathsize 14px style 0 over 0 end style dapat diselesaikan dengan metode pemfaktoran sebagai berikut

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x squared plus 4 x minus 21 over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator left parenthesis x plus 7 right parenthesis left parenthesis x minus 3 right parenthesis over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of x plus 7 end cell row blank equals cell 3 plus 7 end cell row blank equals 10 end table end style

Dengan demikian, limx3x3x2+4x21=10

Oleh karena itu, jawaban yang tepat adalah C.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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