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Selesaikan setiap masalah limit di bawah ini secara intuitif.

Pertanyaan

Selesaikan setiap masalah limit di bawah ini secara intuitif. 

begin mathsize 14px style 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 style

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. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Sibuea

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

...

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 

 

0

Roboguru

JIka maka nilai dari  adalah ...

Pembahasan Soal:

limit as x rightwards arrow 4 of f open parentheses x close parentheses equals limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator 2 square root of x minus x end fraction

Subtitusikan x equals 4 ke persamaan limit tersebut, sehingga didapat sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator 2 square root of x minus x end fraction end cell equals cell fraction numerator 4 minus 4 over denominator 2 square root of 4 minus 4 end fraction end cell row blank equals cell fraction numerator 0 over denominator 2 times 2 minus 4 end fraction end cell row blank equals cell 0 over 0 end cell end table 

Karena menghasilkan limit bentuk tak tentu, maka 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 4 of fraction numerator x minus 4 over denominator 2 square root of x minus x end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator up diagonal strike open parentheses square root of x minus 2 close parentheses end strike open parentheses square root of x plus 2 close parentheses over denominator negative square root of x up diagonal strike open parentheses square root of x minus 2 close parentheses end strike end fraction end cell row blank equals cell fraction numerator square root of 4 plus 2 over denominator negative square root of 4 end fraction end cell row blank equals cell fraction numerator 2 plus 2 over denominator negative 2 end fraction end cell row blank equals cell negative 2 end cell end table 

Jadi, Nilai limit as x rightwards arrow 4 of f open parentheses x close parentheses adalah negative 2

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

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 5 of space fraction numerator x cubed minus 125 over denominator x squared minus 25 end fraction end cell equals cell limit as x rightwards arrow 5 of space fraction numerator open parentheses x minus 5 close parentheses open parentheses x squared plus 5 x plus 25 close parentheses over denominator x squared minus 25 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of space fraction numerator down diagonal strike open parentheses x minus 5 close parentheses end strike open parentheses x squared plus 5 x plus 25 close parentheses over denominator down diagonal strike open parentheses x minus 5 close parentheses end strike open parentheses x plus 5 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 5 of space fraction numerator x squared plus 5 x plus 25 over denominator x plus 5 end fraction end cell end table end style

Substitusi begin mathsize 14px style x end style dengan begin mathsize 14px style 5 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 5 of space fraction numerator x squared plus 5 x plus 25 over denominator x plus 5 end fraction end cell equals cell fraction numerator 5 squared plus 5 open parentheses 5 close parentheses plus 25 over denominator 5 plus 5 end fraction end cell row blank equals cell fraction numerator 25 plus 25 plus 25 over denominator 10 end fraction end cell row blank equals cell 75 over 10 end cell row blank equals cell 15 over 2 end cell end table end style

Maka, nilai dari begin mathsize 14px style limit as x rightwards arrow 5 of space fraction numerator x cubed minus 125 over denominator x squared minus 25 end fraction equals 15 over 2 end style.

0

Roboguru

Pembahasan Soal:

limit as x rightwards arrow 2 of fraction numerator x squared plus 5 x minus 14 over denominator x minus 2 end fraction

Subtitusikan x equals negative 2 ke persamaan limit tersebut, sehingga di dapat sebagai berikut.

limit as x rightwards arrow 2 of fraction numerator x squared plus 5 x minus 14 over denominator x minus 2 end fraction equals fraction numerator 2 squared plus 5 open parentheses 2 close parentheses minus 14 over denominator 2 minus 2 end fraction equals 0 over 0 

Karena menghasilkan limit bentuk tak tentu, maka 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 plus 5 x minus 14 over denominator x minus 2 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator up diagonal strike open parentheses x minus 2 close parentheses end strike open parentheses x plus 7 close parentheses over denominator up diagonal strike x minus 2 end strike end fraction end cell row blank equals cell 2 plus 7 end cell row blank equals 9 end table 

Jadi, Nilai limit as x rightwards arrow 2 of fraction numerator x squared plus 5 x minus 14 over denominator x minus 2 end fraction adalah 9

0

Roboguru

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